10.8 Answers to Chapter Questions

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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.

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