Decimal Arithmetic

Section 1.5 contained a brief introduction to the integer representation systems known as binary coded decimal (BCD). BCD representations are especially useful for storing integers with many digits, such as might be needed for financial records. BCD values are easier than 2's complement values to convert to or from ASCII format, but only a few 80x86 instructions are available to facilitate arithmetic with BCD numbers.

Chapter 11 describes BCD representation schemes and the 80x86 instructions that are used with BCD numbers. It includes code to convert BCD representations for numbers to and from corresponding ASCII representations and some procedures for BCD arithmetic.

Packed BCD Representations

The two major classifications of BCD schemes are packed and unpacked, and many variations with respect to the number of bytes used and how the sign of a value is represented. This section and Section 11.2 discuss packed BCD numbers. Section 11.3 tells about unpacked BCD numbers.

Packed BCD representations store two decimal digits per byte, one in the high-order four bits and one in the low-order four bits. For example, the bit pattern 01101001 represents the decimal number 69, using 0110 for 6 and 1001 for 9. One confusing thing about packed BCD is that this same bit pattern is written 69 in hexadecimal; however, this just means that if 01101001 is thought of as a BCD number, it represents the decimal value 69, but if it is viewed as a signed or unsigned binary integer, the corresponding decimal value is 105. This again makes the point that a given pattern of bits can have multiple numeric interpretations, as well as nonnumeric meanings.

If single bytes were used for packed BCD representations, then decimal numbers from 0 to 99 could be stored. This would not be very useful, so typically several bytes are used to store a single number. Many schemes are possible; some use a fixed number of bytes and some have variable length, incorporating a field for length as part of the representation. The bit pattern for a number often includes one or more bits to indicate the sign of the number.

As mentioned in Chapter 10, the Microsoft Macro Assembler provides a DT directive that can be used to define a 10 byte packed decimal number. Although other representation systems are equally valid, this book concentrates on this scheme. The directive

 DT 123456789

reserves ten bytes of storage with initial values (in hex)

 89 67 45 23 01 00 00 00 00 00

Notice that the bytes are stored backward, low order to high order, but within each byte the individual decimal digits are stored forward. This is consistent with the way that high-order and low-order bytes are reversed in 2's complement integers. The tenth byte in this representation is used to indicate the sign of the entire number. This byte is 00 for a positive number and 80 for a negative number. Therefore the DT directive

 DT -1469

produces

 69 14 00 00 00 00 00 00 00 80

Notice that only the sign indicator changes for a negative number; other digits of the representation are the same as they would be for the corresponding positive number.

Since an entire byte is used for the sign indicator, only nine bytes remain to store decimal digits. Therefore the packed BCD scheme used by the DT directive stores a signed number up to decimal 18 digits long. With MASM 6.11, extra digits are truncated without warning.

Although DT directives can be used to initialize packed BCD numbers in an assembly language program and arithmetic can be done on these numbers with the aid of the instructions covered in the next section, packed BCD numbers are of little service unless they can be displayed for human use. Figure 11.1 gives the source code for a procedure ptoaProc that converts a packed BCD number to the corresponding ASCII string. This procedure does the same job for packed BCD numbers as itoaProc and dtoaProc do for 2's complement integers.

 ptoaProc PROC NEAR
 ; convert 10-byte BCD number to a 19-byte-long ASCII string
 ; parameter 1: address of BCD number
 ; parameter 2: destination address
 ; author: R. Detmer revised: 5/98
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 push eax
 push ecx
 mov esi, [ebp+12] ; source address
 mov edi, [ebp+8] ; destination address
 add edi, 18 ; point to last byte of destination
 mov ecx, 9 ; count of bytes to process
 for1: mov al, [esi] ; byte with two bcd digits
 mov ah, al ; copy to high-order byte of AX
 and al, 00001111b ; mask out higher-order digit
 or al, 30h ; convert to ASCII character
 mov [edi], al ; save lower-order digit
 dec edi ; point at next destination byte to left
 shr ah, 4 ; shift out lower-order digit
 or ah, 30h ; convert to ASCII
 mov [edi], ah ; save higher-order digit
 dec edi ; point at next destination byte to left
 inc esi ; point at next source byte
 loop for1 ; continue for 9 bytes


 mov BYTE PTR [edi], ' ' ; space for positive number
 and BYTE PTR [esi], 80h ; check sign byte
 jz nonNeg ; skip if not negative
 mov BYTE PTR [edi], '-' ; minus sign
 nonNeg:
 pop ecx ; restore registers
 pop eax
 pop esi
 pop edi
 pop ebp
 ret 8 ; return, removing parameters
 ptoaProc ENDP


Figure 11.1: Packed BCD to ASCII conversion

The procedure ptoaProc has two parameters: a 10-byte-long packed BCD source and a 19-byte-long ASCII destination string, each passed by location. The destination is 19 bytes long to allow for a sign and 18 digits. The sign will be a space for a positive number and a minus sign for a negative number. For the digits, leading zeros rather than spaces are produced. The procedure implements the following design:

copy source address to ESI;
copy destination address to EDI;
add 18 to EDI to point at last byte of destination string;

for count := 9 down to 1 loop { process byte containing two digits }
 copy next source byte to AL;
 duplicate source byte in AH;
 mask out high-order digit in AL;
 convert low-order digit in AL to ASCII code;
 store ASCII code for low-order digit in destination string;
 decrement EDI to point at next destination byte to left;
 shift AH 4 bits to right to get only high-order digit;
 convert high-order digit in AH to ASCII code;
 store ASCII code for high-order digit in destination string;
 decrement EDI to point at next destination byte to left;
 increment ESI to point at next source digit to right;
end for;
move space to first byte of destination string;
if source number is negative
then
 move minus sign to first byte of destination string;
end if;

The most interesting part of the design and code is the portion that splits a single source byte into two destination bytes. Two copies of the source byte are made, one in AL and one in AH. The byte in AL is converted to the ASCII code for the low-order digit using an and instruction to mask the left four bits and an or instruction to put 0011 (hex 3) in their place. The high-order digit is processed similarly. A shr instruction discards the low-order digit in AH, moves the high-order digit to the right four bits and zeros the left four bits. Another or instruction produces the ASCII code for the high-order digit.

Once a packed BCD number is converted to an ASCII string it can be displayed using the output macro or by some other means. Since BCD numbers are often used for financial calculations, some other ASCII representation may be more desirable than that generated by ptoaProc. Some exercises at the end of this section specify alternatives.

Sometimes it is necessary to convert an ASCII string to a corresponding packed BCD value. Figure 11.2 shows a procedure atopProc that accomplishes this task in a restricted setting. The procedure has two parameters, the addresses of an ASCII source string and a 10 byte BCD destination string. The ASCII source string is very limited. It can consist only of ASCII codes for digits terminated by a null byte; no sign, no space, nor any other character code is permitted.

atopProc PROC NEAR32
; Convert ASCII string at to 10-byte BCD number
; parameter 1: ASCII string address parameter 2: BCD number address
; null-terminated source string consists only of ASCII codes for digits,
; author: R. Detmer revised: 5/98
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 push eax
 push ecx
 mov esi, [ebp+12] ; source address
 mov edi, [ebp+8] ; destination address
 mov DWORD PTR [edi], 0 ; zero BCD destination
 mov DWORD PTR [edi+4], 0
 mov WORD PTR [edi+8], 0
; find length of source string and move ESI to trailing null
 mov ecx, 0 ; count := 0
while1: cmp BYTE PTR [esi], 0 ; while not end of string (null)
 jz endwhile1
 inc ecx ; add 1 to count of characters
 inc esi ; point at next character
 jmp while1 ; check again
endwhile1:
; process source characters a pair at a time
while2: cmp ecx, 0 ; while count > 0
 jz endwhile2
 dec esi ; point at next ASCII byte from right
 mov al, BYTE PTR [esi] ; get byte
 and al, 00001111b ; convert to BCD digit
 mov BYTE PTR [edi], al ; save BCD digit
 dec ecx ; decrement count
 jz endwhile2 ; exit loop if out of source digits
 dec esi ; point at next ASCII byte from right
 mov al, BYTE PTR [esi] ; get byte
 shl al, 4 ; shift to left and convert to digit
 or BYTE PTR [edi], al ; combine with other BCD digit
 dec ecx ; decrement count
 inc edi ; point at next destination byte
 jmp while2 ; repeat for all source characters
endwhile2:
 pop ecx ; restore registers
 pop eax
 pop esi
 pop edi
 pop ebp
 ret 8 ; return, removing parameters
atopProc ENDP


Figure 11.2: ASCII to packed BCD conversion

The design of procedure atopProc is quite different from atodProc (Fig. 8.9) that produces a doubleword integer from an ASCII string. The ASCII-to-doubleword routine scans source characters left to right one at a time, but the ASCII-to-packed BCD procedure scans the source string right to left, two characters at a time, in order to pack two decimal digits into one byte. The procedure must begin by locating the right end of the string. If there is an odd number of source characters, then only one character will contribute to the last BCD byte. The design for atopProc appears below.

copy source address to ESI;
copy destination address to EDI;
initialize all 10 bytes of destination, each to 00;
counter := 0;
while ESI is not pointing at trailing null byte of ASCII source loop
 add 1 to counter;
 increment ESI to point at next byte of source string;
end while;
while counter > 0 loop
 decrement ESI to point at next source byte from right;
 copy source byte to AL;
 convert ASCII code to digit by zeroing leftmost 4 bits;
 save low-order digit in destination string;
 subtract 1 from counter;
 if counter = 0
 then
 exit loop;
 end if;
 decrement ESI to point at next source byte from right;
 copy source byte to AL;
 shift AL 4 bits left to get digit in high order 4 bits;
 or AL with destination byte to combine with low-order digit;
 subtract 1 from counter;
 increment EDI to point at next destination byte;
end while;

The first while loop in the design simply scans the source string left to right, counting digits preceding the trailing null byte. Although this design allows only ASCII codes for digits, an extra loop could be included to skip leading blanks and a leading minus or plus (− or +) could be noted. (These and other enhancements are specified in programming exercises.)

The second while loop processes the ASCII codes for digits that have been counted in the first loop. Two digits, if available, must be packed into a single destination byte. At least one source byte is there each time through the loop, so the first is loaded into AL, changed from an ASCII code to a digit, and stored in the destination string. (An alternative way to convert the ASCII code to a digit would be to subtract 3016.) If source characters are exhausted, then the while loop is exited. Otherwise a second ASCII character is loaded into AL, a left shift instruction converts it to a digit in the left four bits of AL, and an or combines it with the right digit already stored in memory in the destination string.

The atopProc procedure could be used to convert a string obtained from the input macro. If some other method were used, one would have to ensure that the string has a trailing null byte.

Exercises 11.1

  1. Find the initial values that MASM will generate for each DT directive below:

    1. DT 123456
    2. DT -123456
    3. DT 345
    4. DT -345
    5. DT 102030405060708090
    6. DT -102030405060708090
  2. Explain how you could use floating-point instructions to convert a number stored as a 2's complement doubleword integer to a 10-byte packed decimal equivalent value. From packed BCD to doubleword integer?
  3. Define a macro ptoa similar to the itoa macro described in Section 9.5. Use two parameters, dest and source, dest referencing a 19-byte-long ASCII string and source referencing a 10-byte packed BCD string in memory. Include safeguards to ensure that the correct number of arguments is used in a call. Code in the macro will call ptoaProc.

Programming Exercises 11.1

  1. Modify the code for the ptoaProc procedure so that it produces leading spaces instead of zeros, and so that the minus sign, if any, is placed to the immediate left of the first nonzero digit. If the value of the entire number is zero, the units-position (rightmost) zero is not replaced by a space. The total string length will remain 19 characters. The procedure will remove parameters from the stack.
  2. Modify the code for the ptoaProc procedure so that it produces a 22-byte-long ASCII string giving a monetary representation of the source value. Use leading spaces instead of leading zeros (if any) in the first 16 positions. Character 17 is always a decimal point. Characters 18 and 19 are always digits, even if they have value zero. Character 20 is a space. Characters 21 and 22 are ASCII codes for "CR" if the value is positive and "DB" if the value is negative. The procedure will remove parameters from the stack.

  3.  

    1. Modify the code for the atopProc procedure so that it will skip leading spaces in the source string, accept a leading plus or minus (+ or −) immediately before the first digit, and terminate scanning when any nondigit (rather than only a null byte) is encountered in the string. If a minus sign is encountered, the sign byte of the BCD representation is set to 8016. The procedure will remove parameters from the stack.
    2. Define a macro atop similar to the atoi macro described in Section 9.5. Use two parameters, dest and source, dest referencing a 10-byte packed BCD string in memory and source referencing a 19-byte-long ASCII string. Include safeguards to ensure that the correct number of arguments is used in a call. Code in the macro will call the modified atopProc from part (a).
  4. Write a procedure editProc that has two parameters, (1) the address of a pattern string and (2) the address of a 10-byte packed BCD value. The procedure selectively replaces some characters in the pattern string by spaces or by ASCII codes for digits extracted from the BCD value. Except for a terminating null byte, the only allowable characters in a pattern string are a pound sign (#), a comma (,) and a period (.). A period is always unchanged. Each # is replaced by a digit. There will be at most 18 pound signs and if there are fewer than 18, then lower-order digits from the BCD value are used. Leading zeros in the resulting string are changed to spaces unless they follow a period, in which case they remain zeros. A comma is unchanged unless it is adjacent to a space; such a comma is changed to a space. The following examples (with b indicating a space) illustrate how editProc works. Note that the original pattern is destroyed by the procedure. The procedure will remove parameters from the stack.

    Before pattern

    BCD value

    After pattern

    ##,###.##

    123456

    b1,234.56

    ##,###.##

    12345

    bbb123.45

    ##,###.##

    1

    bbbbbb.01

Packed BCD Instructions

Addition and subtraction operations for packed BCD numbers are similar to those for multicomponent 2's complement numbers (Section 4.5). Corresponding bytes of the two operands are added, and the carry-from-one addition is added to the next pair of bytes. BCD operands have no special addition instruction; the regular add and adc instructions are used. However, these instructions are designed for binary values, not BCD values, so for many operands they give the wrong sums.

The 80x86 architecture includes a daa (decimal adjust after addition) instruction used after an addition instruction to correct the sum. This section explains the operation of the daa instruction and its counterpart das for subtraction. Procedures for addition and subtraction of non-negative 10-byte packed BCD numbers are developed; then a general addition procedure is given.

A few examples illustrate the problem with using binary addition for BCD operands. The AF column gives the value of the auxiliary carry flag, the significance of which is discussed below.

Before

After add al,bl

AL

BL

AL

AF

CF

34

25

59

0

0

37

25

5C

0

0

93

25

B8

0

0

28

39

61

1

0

79

99

12

1

1

Although each answer is correct as the sum of two unsigned binary integers, only the first result is correct as a BCD value. The second and third sums contain bit patterns that are not used in BCD representations, C16 in the second example and B16 in the third. The last two sums contain no invalid digit-they are simply wrong as decimal sums.

The daa instruction is used after an addition instruction to convert a binary sum into a packed BCD sum. The instruction has no operand; the sum to be converted must be in the AL register. A daa instruction examines and sets both the carry flag CF and the auxiliary carry flag AF (bit 4 of the EFLAGS register). Recall that the carry flag is set to 1 during addition of two eight bit numbers if there is a carry out of the leftmost position. The AF flag similarly is set to 1 by add or adc instructions if there is a carry resulting from addition of the low-order four bits of the two operands. One way of thinking of this is that the sum of the two low-order hex digits is greater than F16.

A daa instruction first examines the right hex digit of the binary sum in AL. If this digit is over 9 (that is, A through F), then 6 is added to the entire sum and AF is set to 1. Notice that this would correct the result in the second example above since 5C + 6 = 62, the correct packed BCD sum of 37 and 25. The same correction is applied if AF=1 when the daa instruction is executed. Thus in the fourth example, 61 + 6 = 67.

After correcting the right digit, daa examines the left digit in AL. The action is similar: If the left digit is over 9 or CF=1, then 6016 is added to the entire sum. The carry flag CF is set to 1 if this correction is applied. In the third example, B8 + 60 = 18 with a carry of 1.

Both digits must be corrected in the last example, 12 + 6 = 18 and 18 + 60 = 78 (since CF=1). The chart below completes the above examples, assuming that both of the following instructions are executed.

add al, bl
daa

Before

After add

After daa

AL: 34

AL: 59

AL: 59

BL: 25

AF: 0 CF: 0

AF: 0 CF: 0

AL: 37

AL: 5C

AL: 62

BL: 25

AF: 0 CF: 0

AF: 1 CF: 0

AL: 93

AL: B8

AL: 18

BL: 25

AF: 0 CF: 0

AF: 0 CF: 1

AL: 28

AL: 61

AL: 67

BL: 39

AF: 1 CF: 0

AF: 1 CF: 0

AL: 79

AL: 12

AL: 78

BL: 99

AF: 1 CF: 1

AF: 1 CF: 1

The das instruction (decimal adjust after subtraction) is used after a sub or sbb instruction. It acts like the daa except that 6 or 6016 is subtracted from rather than added to the value in AL. The following examples show how das works following sub al,bl. In the first example, both CF and AF are set to 1 since the subtraction requires borrows in both digit positions. When 6 and 6016 are subtracted from BC, the result is 56, and both CF and AF remain set to 1. This is the correct answer since 25 - 69 = 56 (borrowing 1 to change 25 into 125.)

Before

After sub

After das

AL: 25

AL: BC

AL: 56

BL: 69

AF: 1 CF: 1

AF: 1 CF: 1

AL: 37

AL: 12

AL: 12

BL: 25

AF: 0 CF: 0

AF: 0 CF: 0

AL: 93

AL: 6E

AL: 68

BL: 25

AF: 1 CF: 0

AF: 1 CF: 1

AL: 92

AL: 59

AL: 53

BL: 39

AF: 1 CF: 0

AF: 1 CF: 0

AL: 79

AL: E4

AL: 84

BL: 95

AF: 0 CF: 1

AF: 0 CF: 1

Each of the daa and das instructions encodes in a single byte. The daa instruction has opcode 27 and the das instruction has opcode 2F. Each requires three clock cycles to execute on a Pentium. In addition to modifying AF and CF, the SF, ZF and PF flags are set or reset by daa or das instructions to correspond to the final value in AL. The overflow flag OF is undefined and other flags are not affected.

The first BCD arithmetic procedure in this section adds two non-negative 10-byte numbers. This procedure will have two parameters, addresses of destination and source values, respectively. Each will serve as an operand, and the destination will be replaced by the sum, consistent with the way that ordinary addition instructions use the destination operand. We will not be concerned about setting flags; the exercises specify a more complete procedure that assigns appropriate values to SF, ZF, and CF. A design for the procedure is given below. This design is implemented in the procedure addBcd1 (see Fig. 11.3).

 addBcd1 PROC NEAR32
 ; add two non-negative 10 byte packed BCD numbers
 ; parameter1: address of operand1 (and destination)
 ; parameter2: address of operand2
 ; author: R. Detmer revised: 5/98
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 push ecx
 push eax
 mov edi, [ebp+12] ; destination address
 mov esi, [ebp+8] ; source address


 clc ; clear carry flag for first add
 mov ecx, 9 ; count of bytes to process
 forAdd: mov al, [edi] ; get one operand byte
 adc al, [esi] ; add other operand byte
 daa ; adjust to BCD
 mov [edi], al ; save sum
 inc edi ; point at next operand bytes
 inc esi
 loop forAdd ; repeat for all 9 bytes


 pop eax ; restore registers
 pop ecx
 pop edi
 pop esi
 pop ebp
 ret 8 ; return to caller


 addBcd1 ENDP


Figure 11.3: Addition of non-negative packed BCD numbers

point at first source and destination bytes;
for count := 1 to 9 loop
 copy destination byte to AL;
 add source byte to AL;
 use daa to convert sum to BCD;
 save AL in destination;
 point at next source and destination bytes;
end for;

A subtraction procedure for 10-byte packed BCD numbers is more difficult. Even with the operands restricted to non-negative values, subtracting the source value (address in parameter 2) from the destination (address in parameter 1) will produce a negative result if the source is larger than the destination. A design for the procedure is below.

point at first source and destination bytes;
for count := 1 to 9 loop
 copy destination byte to AL;
 subtract source byte from AL;
 use das to convert difference to BCD;
 save AL in destination string;
 point at next source and destination bytes;
end for;

if source > destination
then
 point at first destination byte;
 for count := 1 to 9 loop
 put 0 in AL;
 subtract destination byte from AL;
 use das to convert difference to BCD;
 save AL in destination string;
 increment DI;
 end for;
 move sign byte 80 to destination string;
end if;

The first part of this design is almost the same as the design for addition. The condition (source > destination) is true if the carry flag is set after the first loop, and the difference is corrected by subtracting it from zero. If this were not done, then, for example, 3 −7 would produce 999999999999999996 instead of −4. This design is implemented as procedure subBcd1 in Fig. 11.4.

 subBcd1 PROC NEAR32
 ; subtract 2 non-negative 10 byte packed BCD numbers
 ; parameter1: address of operand1 (and destination)
 ; parameter2: address of operand2
 ; operand1 -- operand2 stored at destination
 ; author: R. Detmer revised: 5/98
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 push ecx
 push eax
 mov edi, [ebp+12] ; destination address (operand 1)
 mov esi, [ebp+8] ; source address (operand 2)
 clc ; clear carry flag
 mov ecx, 9 ; count of bytes to process
 forSub: mov al, [edi] ; get one operand byte
 sbb al, [esi] ; subtract other operand byte
 das ; adjust to BCD
 mov [edi], al ; save difference
 inc edi ; point at next operand bytes
 inc esi
 loop forSub ; repeat for all 9 bytes


 jnc endIfBigger ; done if destination >= source
 sub edi, 9 ; point at beginning of destination
 mov ecx, 9 ; count of bytes to process
 forSub1: mov al,0 ;subtract destination from zero
 sbb al, [edi]
 das
 mov [edi], al
 inc edi ; next byte
 loop forSub1
 mov BYTE PTR [edi], 80h ; negative result
 endIfBigger:
 pop eax ; restore registers
 pop ecx
 pop edi
 pop esi
 pop ebp
 ret 8 ; return to caller
 subBcd1 ENDP


Figure 11.4: Subtraction of non-negative packed BCD numbers

Once you have the addBcd1 and subBcd1 procedures that combine non-negative operands, it is not too difficult to construct the general packed BCD addition and subtraction procedures. The design for addition is

if operand1 ≥ 0
then
 if operand2 ≥ 0
 then
 addBcd1(operand1, operand2);
 else
 subBcd1(operand1, operand2);
 end if;
else {operand1 < 0}
 if (operand2 < 0)
 then
 addBcd1(operand1, operand2);
 else
 change sign byte of operand1;
 subBcd1(operand1, operand2);
 change sign byte of operand1;
 end if;
end if;

The design for negative operand1 is a little tricky. When operand2 is also negative, the result will be negative. Since addBcd1 does not affect the sign byte of the destination (operand1), the result after adding operand2 will be negative with no special adjustment required. Adding a non-negative operand2 can result in either a positive or negative result. The reader should verify that this design and corresponding code produces the correct sign for the result. This design is implemented in procedure addBcd, shown in Fig. 11.5. A general procedure for subtraction is left as an exercise.

 addBcd PROC NEAR32
 ; add two arbitrary 10 byte packed BCD numbers
 ; parameter1: address of operand1 (and destination)
 ; parameter2: address of operand2
 ; author: R. Detmer revised: 5/98
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 mov edi, [ebp+12] ; destination address
 mov esi, [ebp+8] ; source address
 push edi ; parameter1 for next call
 push esi ; parameter2 for next call
 cmp BYTE PTR [edi+9], 80h ; operand1 >= 0?
 je op1Neg
 cmp BYTE PTR [esi+9], 80h ; operand2 >= 0?
 je op2Neg
 call addBcd1 ; add (>=0, >=0)
 jmp endIfOp2Pos
 op2Neg: call subBcd1 ; sub (>=0, <0)
 endIfOp2Pos:
 jmp endIfOp1Pos ; done
 op1Neg: cmp BYTE PTR [esi+9], 80h ; operand2 < 0 ?
 jne op2Pos
 call addBcd1 ; add (<0, <0)
 jmp endIfOp2Neg
 op2Pos: xor BYTE PTR [edi+9], 80h ; change sign byte
 call subBcd1 ; sub (<0, >=0)
 xor BYTE PTR [edi+9], 80h ; change sign byte
 endIfOp2Neg:
 endIfOp1Pos:
 pop edi ; restore registers
 pop esi
 pop ebp
 ret 8 ; return to caller


 addBcd ENDP


Figure 11.5: General BCD addition procedure

Exercises 11.2

  1. In each part below, assume that the instructions

    add al, bl
    daa
    

    are executed. Give the values in the AL register, carry flag CF, and auxiliary flag AF: (1) after the add and before the daa and (2) after the daa.

    1. AL: 35 BL: 42
    2. AL: 27 BL: 61
    3. AL: 35 BL: 48
    4. AL: 47 BL: 61
    5. AL: 35 BL: 92
    6. AL: 27 BL: 69
    7. AL: 75 BL: 46
    8. AL: 00 BL: 61
    9. AL: 85 BL: 82
    10. AL: 89 BL: 98
    11. AL: 76 BL: 89
    12. AL: 27 BL: 00
  2. Repeat the parts of Exercise 1 for the instructions

    sub al, bl
    das
    

Programming Exercises 11.2

  1. Modify the addBcd procedure to set SF, ZF, and CF. The sign flag will be set according to the sign of the sum, and ZF will be set for a zero result. The carry flag CF will be set if there are more than 18 digits in the sum.
  2. Design and code a general subtraction procedure subBcd with two parameters: (1) the address of operand1 and (2) the address of operand2. The difference operand1-operand2 will be stored at the address of operand1. The procedure will remove parameters from the stack.

Unpacked BCD Representations and Instructions

Unpacked BCD numbers differ from packed representations by storing one decimal digit per byte instead of two. The bit pattern in the left half of each byte is 0000. This section describes how to define unpacked BCD numbers, how to convert this representation to and from ASCII, and how to use 80x86 instructions to do some arithmetic operations with unpacked BCD numbers.

Unpacked BCD representations have no standard length. In this book each value will be stored in eight bytes, with high-order digits on the left and low-order digits on the right (opposite to the way a DT directive stores packed BCD numbers). No sign byte will be used, so only non-negative numbers will be represented. An ordinary BYTE directive can be used to initialize an unpacked BCD value. For example, the statement

 BYTE 0,0,0,5,4,3,2,8

reserves eight bytes of storage containing 00 00 00 05 04 03 02 08, the unpacked BCD representation for 54328. The directive

 BYTE 8 DUP (?)

establishes an eight-byte-long area that can be used to store an unpacked BCD value.

It is simple to convert an unpacked BCD value to or from ASCII. Suppose that the data segment of a program includes the directives

 ascii DB 8 DUP (?)
 unpacked DB 8 DUP (?)

If unpacked already contains an unpacked BCD value, the following code fragment will produce the corresponding ASCII representation at ascii.

 lea edi, ascii ; destination
 lea esi, unpacked ; source
 mov ecx, 8 ; bytes to process
 for8: mov al, [esi] ; get digit
 or al, 30h ; convert to ASCII
 mov [edi], al ; store ASCII character
 inc edi ; increment pointers
 inc esi
 loop for8 ; repeat for all bytes

Converting from an ASCII string to an unpacked BCD representation is equally easy. The same loop structure can be used with the roles of EDI and ESI reversed, and with the or instruction replaced by

 and al, 0fh ; convert ASCII to unpacked BCD

to mask the high-order four bits. Conversions between ASCII and unpacked BCD are even simpler if they are done "in place" (see Exercise 3).

The 80x86 architecture includes four instructions to facilitate arithmetic with unpacked BCD representations. Each mnemonic begins with "aa" for "ASCII adjust"— Intel uses the word ASCII to describe unpacked BCD representations, even though the ASCII representation for a digit has 0011 in the left half byte and the unpacked representation has 0000. The four instructions are aaa, aas, aam, and aad. Information about these instructions is given in Fig. 11.6.

Instruction

Mnemonic

Number of bytes

Opcode

Clocks (Pentium)

ASCII adjust after addition

aaa

1

37

3

ASCII adjust for subtraction

aas

1

3F

3

ASCII adjust after multiplication

aam

2

D4 0A

18

ASCII adjust before division

aad

2

D5 0A

10


Figure 11.6: Unpacked BCD instructions

The aaa and aas instructions are similar to their packed BCD counterparts daa and das. For addition, bytes containing unpacked BCD operands are combined using an add or adc instruction, yielding a sum in the AL register. An aaa instruction then corrects the value in AL if necessary. An aaa instruction sets flags and may also affect AH; recall that a daa affects only AL and flags. The following algorithm describes how aaa works.

if (right digit in AL > 9) or (AF=1)
then
 add 6 to AL;
 increment AH;
 AF := 1;
end if;

CF := AF;
left digit in AL := 0;

The action of an aas instruction is similar. The first two operations inside the if are replaced by

 subtract 6 from AL;
 decrement AH;

The OF, PF, SF, and ZF flags are left undefined by aaa and aas instructions.

Here are some examples of showing how add and aaa work together. In each example, assume that the following pair of instructions is executed.

 add al, ch
 aaa

Before

After add

After aaa

AX: 00 04

AX: 00 07

AX: 00 07

CH: 03

AF: 0

AF: 0 CF: 0

AX: 00 04

AX: 00 0B

AX: 01 01

CH: 07

AF: 0

AF: 1 CF: 1

AX: 00 08

AX: 00 11

AX: 01 07

CH: 09

AF: 1

AF: 1 CF: 1

AX: 05 05

AX: 05 0C

AX: 06 02

CH: 07

AF: 0

AF: 1 CF: 1

Another set of examples illustrates how sub and aas find differences of single byte unpacked BCD operands. This time assume that the following instructions are executed.

 sub al, dl
 aas

Before

After sub

After aas

AX: 00 08

AX: 00 05

AX: 00 05

DL: 03

AF: 0

AF: 0 CF: 0

AX: 00 03

AX: 00 FC

AX: FF 06

DL: 07

AF: 1

AF: 1 CF: 1

AX: 05 02

AX: 05 F9

AX: 04 03

DL: 09

AF: 1

AF: 1 CF: 1

Figure 11.7 displays a procedure addUnp that adds two eight-byte unpacked BCD numbers whose addresses are passed as parameters. This procedure is simpler than the similar addBcd1 procedure in Fig. 11.3. No effort is made to produce significant flag values. Since low-order digits are stored to the right, the bytes are processed right to left. (Programming Exercise 1 specifies the corresponding procedure for subtraction.)

 addUnp PROC NEAR32
 ; add two 8-byte unpacked BCD numbers
 ; parameter 1: operand1 and destination address
 ; parameter 2: operand2 address
 ; author: R. Detmer revised: 5/98
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 push eax
 push ecx
 mov edi, [ebp+12] ; destination address
 mov esi, [ebp+8] ; source address
 add esi, 8 ; point at byte after source
 add edi, 8 ; byte after destination
 clc ; clear carry flag
 mov ecx, 8 ; count of bytes to process
 forAdd: dec edi ; point at operand bytes to left
 dec esi
 mov al, [edi] ; get one operand byte
 adc al, [esi] ; add other operand byte
 aaa ; adjust to unpacked BCD
 mov [edi], al ; save sum
 loop forAdd ; repeat for all 8 bytes
 pop ecx ; restore registers
 pop eax
 pop edi
 pop esi
 pop ebp
 ret 8 ; return, discarding paramters
 addUnp ENDP


Figure 11.7: Addition of two 8-byte unpacked BCD numbers

One interesting feature of the procedure addUnp is that it will give the correct unpacked BCD sum of eight byte ASCII (not unpacked BCD) numbers—Intel’s use of "ASCII" in the unpacked BCD mnemonics is not as unreasonable as it first seems. The procedure is successful for ASCII strings since the action of the aaa instruction depends only on what add does with low-order digits, and aaa always sets the high-order digit in AL to zero. However, even if the operands are true ASCII character strings, the sum is not ASCII; it is unpacked BCD.

Two single byte unpacked BCD operands are multiplied using an ordinary mul instruction, resulting in a product in the AX register. Of course, this product will be correct as a binary number, not usually as a BCD value. The aam instruction converts the product in AX to two unpacked BCD digits in AH and AL. In effect, an aam instruction divides the number in AL by 10, putting the quotient in AH and the remainder in AL. The following examples assume that the instructions

 mul bh
 aam

are executed.

Before

After mul

After aam

AX: 00 09

AX: 00 51

AX: 08 01

BH: 06

   

AX: 00 05

AX: 00 1E

AX: 03 00

BH: 06

   

AX: 00 06

AX: 00 2A

AX: 04 02

BH: 07

   

Some flags are affected by an aam instruction. The PF, SF, and ZF flags are given values corresponding to the final value in AX; the AF, CF, and OF flags are undefined.

Multiplication of single-digit numbers is not very useful. Figure 11.8 gives a procedure mulUnp1 to multiply an eight-byte unpacked BCD number by a single-digit unpacked BCD number. The procedure has three parameters: (1) the destination address, (2) the address of the BCD source, and (3) a word containing the single-digit unpacked BCD number as its low-order byte.

 mulUnp1 PROC NEAR32
 ; multiply 8 byte and 1 byte unpacked BCD numbers
 ; parameter 1: destination address
 ; parameter 2: address of 8 byte unpacked BCD number
 ; parameter 3: word w/ low-order byte containing 1-digit BCD nbr
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 push eax
 push ebx
 push ecx
 mov edi, [ebp+14] ; destination address
 mov esi, [ebp+10] ; source address
 mov bx, [ebp+8] ; multiplier
 add esi, 8 ; point at byte after source
 add edi, 8 ; byte after destination
 mov bh, 0 ; lastCarry := 0
 mov ecx, 8 ; count of bytes to process
 forMul: dec esi ; point at operand byte to left
 dec edi ; and at destination byte
 mov al, [esi] ; digit from 8 byte number
 mul bl ; multiply by single byte
 aam ; adjust to unpacked BCD
 add al, bh ; add lastCarry
 aaa ; adjust to unpacked BCD
 mov [edi], al ; store units digit
 mov bh, ah ; store lastCarry
 loop forMul ; repeat for all 8 bytes
 pop ecx ; restore registers
 pop ebx
 pop eax
 pop edi
 pop esi
 pop ebp
 ret 10 ; return, discarding paramters
 mulUnp1 ENDP


Figure 11.8: Multiplication of unpacked BCD numbers

The algorithm implemented is essentially the same one as used by grade school children. The single digit is multiplied times the low-order digit of the multi-digit number, the units digit is stored, and the tens digit is recorded as a carry to add to the next product. All eight products can be treated the same by initializing a last-Carry variable to zero prior to beginning a loop. Here is the design that is actually implemented.

{ multiply X7X6X5X4X3X2X1X0 times Y giving Z7Z6Z5Z4Z3Z2Z1Z0}
lastCarry := 0;
for i := 0 to 7 loop
 multiply Xi times Y;
 add lastCarry;
 Zi := units digit;
 lastCarry := tens digit;
end for;

In the code for mulUnp1, the value for lastCarry is stored in the BH register. After a digit from the eight-byte BCD value is multiplied by the single digit in BL, the product is adjusted to unpacked BCD and lastCarry is added. It is then necessary to adjust the sum to unpacked BCD.

The aad instruction essentially reverses the action of the aam instruction. It combines a two-digit unpacked BCD value in AH and AL into a single binary value in AX, multiplying the digit in AH by 10 and adding the digit in AL. The AH register is always cleared to 00. The PF, SF, and ZF flags are given values corresponding to the result; AF, CF, and OF are undefined.

The aad instruction is used before a div instruction, contrary to the other ASCII adjust instructions that are used after the corresponding arithmetic instructions. The examples below assume that the instructions

 aad
 div dh

are executed.

Before

After aad

After div

AX: 07 05

AX: 00 4B

AX: 03 09

DH: 08

DH: 08

 

AX: 06 02

AX: 00 3E

AX: 02 0F

DH: 04

DH: 04

 

AX: 09 03

AX: 00 5D

AX: 01 2E

DH: 02

DH: 02

 

In the first example, the quotient and remainder are in BCD format in AL and AH, respectively, following the div instruction. However, the second and third examples show that this is not always the case. The remainder is correct in AH because it is a binary remainder following division by a number 9 or smaller. The remainder must be 0 through 8, and for numbers in this range a single byte binary value agrees with the unpacked BCD representation. The quotient in AL is obviously a binary number, not a BCD representation. To convert it to unpacked BCD, an aam instruction needs to follow the div. In the second example, this would change AX to 01 05, the correct quotient for 62 ÷ 4. In the third example, aam would yield 0406 in AX, again the correct quotient. Notice that the remainder from the division is lost, so if it is needed, it must be copied from AH before aam is executed.

Notice that the problems illustrated by the previous examples cannot occur when the original digit in AH is smaller than the divisor in DH. The elementary school algorithm for dividing a single digit into a multidigit number works left to right through the dividend, dividing a two digit number by the divisor. The first of the two digits is the remainder from the previous division, which must be smaller than the divisor. The following design formalizes the grade school algorithm.

{ divide X7X6X5X4X3X2X1X0 by Y giving Z7Z6Z5Z4Z3Z2Z1Z0 }
lastRemainder := 0;
for i := 7 downto 0 loop
 dividend := 10*lastRemainder + Xi;
 divide dividend by Y getting quotient & lastRemainder;
 Zi := quotient;
end for;

Code that implements this design is given in Fig. 11.9. The AH register is ideally suited to store lastRemainder since that is where the remainder ends up following division of a 16-bit binary number by an 8-bit number.

 divUnp1 PROC NEAR32
 ; parameter 1: destination address
 ; parameter 2: address of 8 byte unpacked BCD number
 ; parameter 3: word w/ 1-digit BCD number as low-order byte
 ; author: R. Detmer revised: 5/98
 push ebp ; establish stack frame
 mov ebp, esp
 push esi ; save registers
 push edi
 push eax
 push ebx
 push ecx
 mov edi, [ebp+14] ; destination address
 mov esi, [ebp+10] ; source address
 mov bx, [ebp+8] ; divisor
 mov ah, 0 ; lastRemainder := 0
 mov ecx, 8 ; count of bytes to process
 forDiv: mov al, [esi] ; digit from 8 byte number
 aad ; adjust to binary
 div bl ; divide by single byte
 mov [edi], al ; store quotient
 inc esi ; point at next digit of dividend
 inc edi ; and at next destination byte
 loop forDiv ; repeat for all 8 bytes
 pop ecx ; restore registers
 pop ebx
 pop eax
 pop edi
 pop esi
 pop ebp
 ret 10 ; return, discarding paramters
 divUnp1 ENDP


Figure 11.9: Division of unpacked BCD numbers

Exercises 11.3

  1. In each part below, assume that the instructions

    add al, bl
    aaa
    

    are executed. Give the values in the AX register, carry flag CF, and auxiliary flag AF: (1) after the add and before the aaa and (2) after the aaa.

    1. AX: 00 05 BL: 02
    2. AX: 02 06 BL: 03
    3. AX: 03 05 BL: 08
    4. AX: 00 07 BL: 06
    5. AX: 00 09 BL: 08
    6. AX: 02 07 BL: 09
    7. AX: 04 01 BL: 09
    8. AX: 00 00 BL: 01
  2. Repeat the parts of Exercise 1 for the instructions

    sub al, bl
    aas
    
  3. Both parts of this problem assume the definition

    value BYTE 8 DUP(?)
    
    1. Assume that value contains ASCII codes for digits 0 through 9. Write a code fragment to replace these bytes "in place" (without copying bytes to another location) by the corresponding unpacked BCD values.
    2. Assume that value contains an eight-byte-long unpacked BCD value. Write a code fragment to replace these bytes in place by the corresponding ASCII codes for digits 0 through 9.
  4. In each part below, assume that the instructions

    mul ch
    aam
    

    are executed. Give the values in the AX register: (1) after the mul and before the aam and (2) after the aam.

    1. AL: 05 CH: 02
    2. AL: 06 CH: 03
    3. AL: 03 CH: 08
    4. AL: 07 CH: 06
    5. AL: 09 CH: 08
    6. AL: 07 CH: 09
    7. AL: 04 CH: 09
    8. AL: 08 CH: 01
  5. In each part below, assume that the instructions

    aad
    div dl
    aam
    

    are executed. Give the values in the AX register: (1) after the aad and before the div, (2) after the div and before the aam, and (3) after the aam.

    1. AX: 07 05 DL: 08
    2. AX: 05 06 DL: 09
    3. AX: 02 07 DL: 08
    4. AX: 04 07 DL: 06
    5. AX: 05 09 DL: 06
    6. AX: 03 07 DL: 07
    7. AX: 07 04 DL: 03
    8. AX: 05 00 DL: 04

Programming Exercises 11.3

  1. Write a procedure subUnp to find the difference of two eight-byte unpacked BCD numbers. The procedure will have two parameters: (1) the address of operand1 and destination and (2) the address of operand2. The value of operand1 operand2 will be stored at destination. The procedure will set CF to 1 if the source is larger than the destination and clear it to 0 otherwise. Other flag values will not be changed. The procedure will remove parameters from the stack.
  2. Here is one possible variable-length representation for multibyte unpacked BCD numbers. An unsigned binary value in the first byte tells how many decimal digits are in the number. Then digits are stored right to left (low order to high order). For example, the decimal number 1234567890 could be stored 0A 00 09 08 07 06 05 04 03 02 01. This system allows for decimal numbers up to 255 digits long to be stored.

    Write a procedure addVar that adds two unpacked BCD numbers stored in this variable length format. The procedure will have two parameters: (1) the address of operand1 and destination and (2) the address of operand2. The value of operand1 + operand2 will be stored at destination. The two numbers are not necessarily the same length. The sum may be the same length as the longer operand, or one byte longer. Assume that sufficient space has been reserved in the destination field for the sum, even if operand1 is the shorter operand. The procedure will remove parameters from the stack.

Other Architectures VAX Packed Decimal Instructions

Since the 80x86 architecture provides very limited support for packed decimal operations, a large procedure library is necessary to use packed decimal types. Some other architectures provide extensive hardware support for packed decimal. This section briefly examines packed decimal instructions defined in the VAX architecture, although not necessarily implemented in all VAX machines.

The VAX architecture defines a packed decimal string by its length and starting address. The length gives the number of decimal digits stored in the string, not the number of bytes. The last four bits (half byte) are always a sign indicator, normally C16 for positive and D16 for negative. Since decimal digits are packed two per byte, the length (in bytes) of a packed decimal string is approximately half the number of digits. More precisely, for n decimal digits it is (n + 1)/2 if n is odd and (n + 2)/2 if n is even.

The VAX architecture includes a complete set of instructions for performing packed decimal arithmetic: ADDP (add packed), DIVP (divide packed), MULP (multiply packed), and SUBP (subtract packed). Each of these has at least four operands to specify the length and address of each of the packed decimal strings involved. When just two strings are specified, one serves both as a source and the destination. All also have six-operand formats where the sources are specified separately from the destination. (MULP and DIVP have only the six-operand formats.) The MOVP (move packed) instruction copies a packed decimal string from one address to another. The CMPP (compare packed) instruction compares two packed decimal strings, setting condition codes (flags).

Recall the difficulty of converting packed decimal to or from other formats. The VAX architecture provides six different instructions for this purpose. There are functions to convert between packed decimal strings and 32-bit 2's complement integers, and others to convert between packed decimal and numeric strings (including ASCII). There is also an EDIT instruction that converts a packed decimal string to a character string, performing many possible editing operations during the conversions. (Programming Exercises 11.1, #4, describes a similar, but much simpler, editing job.)

The COBOL language directly supports packed decimal types and operations. If you were writing a COBOL compiler for a VAX, then the packed decimal instructions would greatly simplify the job. The resulting compiler would yield much more compact and efficient code than if each packed decimal operation were emulated by a software procedure.

Summary

Integers may be stored in a computer in binary coded decimal form instead of unsigned or 2’s complement binary form. There are two basic BCD systems, packed and unpacked. Packed BCD values store two decimal digits per byte, and unpacked BCD values store a single decimal digit per byte.

Binary representations are much more compact than BCD representations and the 80x86 processor has more instructions for doing arithmetic with binary numbers. However, BCD representations can easily store very large integers and are simple to convert to or from ASCII.

BCD systems may use a variable or a fixed number of bytes and may or may not store a sign indicator. The MASM assembler provides a DT directive that can produce a ten byte signed, packed BCD number. Unpacked BCD numbers can be initialized using BYTE directives.

Arithmetic is done with BCD numbers by combining pairs of bytes from two operands using ordinary binary arithmetic instructions. The binary results are then adjusted to BCD. Packed decimal representations use daa (decimal adjust for addition) and das (decimal adjust for subtraction) instructions. Using these instructions along with binary arithmetic instructions, arithmetic procedures for packed BCD numbers can be developed.

Four instructions are used for unpacked BCD arithmetic: aaa (ASCII adjust for addition), aas (ASCII adjust for subtraction), aam (ASCII adjust for multiplication), and aad (ASCII adjust for division). The aad instruction is different from the others in that it is applied to a BCD result to convert it to binary before applying a div instruction.

Some other architectures provide a much more complete set of packed decimal instructions. In particular, the VAX architecture includes arithmetic, data movement, comparison, and conversion instructions.





Introduction to 80x86 Assembly Language and Computer Architecture
Introduction to 80x86 Assembly Language and Computer Architecture
ISBN: 0763772232
EAN: 2147483647
Year: 2000
Pages: 113
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