Macros argument evaluation in C++

Question

I don't understand what is the reason that macros don't evaluate their argument only once (like inline functions), but each time it is used in the code.

For example:

(the example is taken from here: https://docs.microsoft.com/en-us/cpp/cpp/inline-functions-cpp?view=msvc-170 )

In this code there is a macro:

// inline_functions_macro.c
#include <stdio.h>
#include <conio.h>

#define toupper(a) ((a) >= 'a' && ((a) <= 'z') ? ((a)-('a'-'A')):(a))

int main() {
   char ch;
   printf_s("Enter a character: ");
   ch = toupper( getc(stdin) );
   printf_s( "%c", ch );
}
//  Sample Input:  xyz
// Sample Output:  Z

Compared to this code, with the equivalent inline function:

// inline_functions_inline.cpp
#include <stdio.h>
#include <conio.h>

inline char toupper( char a ) {
   return ((a >= 'a' && a <= 'z') ? a-('a'-'A') : a );
}

int main() {
   printf_s("Enter a character: ");
   char ch = toupper( getc(stdin) );
   printf_s( "%c", ch );
}

Answer 1

Macros are not very smart. They simply perform text substitution before compilation begins.

The code:

#define toupper(a) ((a) >= 'a' && ((a) <= 'z') ? ((a)-('a'-'A')):(a))

int main() {
   char ch = toupper( getc(stdin) );
}

Instructs the preprocessor to change this code and instead hand this to the compiler:

int main() {
   char ch = ((getc(stdin)) >= 'a' && ((getc(stdin)) <= 'z') ? ((getc(stdin))-('a'-'A')):(getc(stdin)));
}

You can observe that toupper was replaced by the defined expression, while every occurance of a is replaced by whatever text was passed as the macro parameter. It's not a value or an expression - it's just text.

Asde from the problems you observe, you may also discover that compiler errors will show this substituted code. Code that you don't see in your source file.