10.5 Advanced Features

10.8 Answers to Chapter Questions

Answer 10-1: After the program has been run through the preprocessor, the std::cout statement is expanded to look like:

std::cout << "The square of all the parts is " <<  7 + 5 * 7 + 5  << '\n';

The equation 7 + 5 * 7 + 5 evaluates to 47. It is a good rule to put parentheses ( ) around all expressions in macros. If you change the definition of ALL_PARTS to:

#define ALL_PARTS (FIRST_PART + LAST_PART)

the program executes correctly.

Answer 10-2: The preprocessor is a very simple-minded program. When it defines a macro, everything past the identifier is part of the macro. In this case, the definition of MAX is literally =10 . When the for statement is expanded, the result is:

for (counter==10; counter > 0; --counter)

C++ allows you to compute a result and throw it away. For this statement, the program checks to see whether counter is 10 and discards the answer. Removing the = from the macro definition will correct the problem.

Answer 10-3: As with the previous problem, the preprocessor does not respect C++ syntax conventions. In this case, the programmer used a semicolon to end the statement, but the preprocessor included it as part of the definition for size . The assignment statement for size , expanded, is:

size = 10; -2;;

The two semicolons at the end do not hurt anything, but the one in the middle is a killer. This line tells C++ to do two things: assign 10 to size and compute the value -2 and throw it away (this results in the null effect warning). Removing the semicolons will fix the problem.

Answer 10-4: The output of the preprocessor looks like:

int main(  ) {      
    int value;    
     
    value = 1;  
    if (value < 0)  
        std::cout << "Fatal Error: Abort\n"; exit(8);  

    std::cout << "We did not die\n"; 
    return (0); 
}

The problem is that two statements follow the if line. Normally they would be put on two lines. If we properly indent this program we get:

Example 10-10. die3/die.cpp

#include <iostream>
#include <cstdlib>

int main(  ) {     
    int value;  // a random value for testing 
    
    value = 1; 
    if (value < 0) 
        std::cout << "Fatal Error:Abort\n";

    exit(8); 

    std::cout << "We did not die\n";
    return (0);
}

From this it is obvious why we always exit. The fact that there were two statements after the if was hidden by using a single preprocessor macro. The cure for this problem is to put curly braces around all multistatement macros.

#define DIE \ 
    {std::cout << "Fatal Error: Abort\n"; exit(8);}

Answer 10-5: The problem is that the preprocessor does not understand C++ syntax. The macro call:

SQR(counter+1)

expands to:

(counter+1 * counter+1)

The result is not the same as ((counter+1) * (counter+1)) . To avoid this problem, use inline functions instead of parameterized macros:

inline int SQR(int x) { return (x*x);}

If you must use parameterized macros, enclose each instance of the parameter in parentheses:

#define SQR(x) ((x) * (x))

Answer 10-6: The only difference between a parameterized macro and one without parameters is the parentheses immediately following the macro name . In this case, a space follows the definition of RECIPROCAL , so it is not a parameterized macro. Instead it is a simple text replacement macro that replaces RECIPROCAL with:

(number) (1.0 / number)

Removing the space between RECIPROCAL and (number) corrects the problem.