Why is the return value not what I expected in this C program with macro? [duplicate]

问题

This question already has answers here:

Strange behaviour of macros C/C++ (5 answers)

Strange behavior of Macro-expansion (3 answers)

Closed 6 years ago.

When I run the following code, the return value is 11, but I was expecting it to return 25. Can someone explain this?

#include<stdio.h>
#define SQR(a) a*a

int main()
{
    int i=3;
    printf("%d",SQR(i+2));
    return 1;
}

回答1:

Needs more parentheses. This:

#define SQR(a) a*a

expands to this:

i+2*i+2

which is:

3+2*3+2

which is 11 because * has precedence over +.

You need to define your macro like this:

#define SQR(a) ((a)*(a))

to ensure that this kind of thing doesn't happen.

回答2:

Macros are not the same as regular functions.

During the reprocessing all macros are replaced exactly by what they are define. In your case, the line:

 printf("%d",SQR(i+2));

is replaced by the line:

 printf("%d", i+2*i+2);

So, you see the unexpected result there.

The correct way is:

#define SQR(a) ((a)*(a))

The preprocessor result will be:

printf("%d", ((i+2)*(i+2)));

Try to learn on this mistake. Issues of this kind are quite hard to debug.

来源：https://stackoverflow.com/questions/18557321/why-is-the-return-value-not-what-i-expected-in-this-c-program-with-macro