I have defined a macro sqr(x) which takes one argument and return x*x. When I call the macro with sqr(5+4), it return 29. Can someone please explain?

Question

Here is the code :

#include<stdio.h>
#define sqr(x) x*x
void main()
{
    printf("\n%d",sqr(9));
    printf("\n%d",sqr(5+4));
}

The output is : 81 29

Answer 1

Having SQR as a macro will never be correct. Make it a function. In C++, the function can be constexpr, in C it should probably be inline.

The following code goes through all iterations of attempting to fix the macro. Note how only the function gives the correct result and no macro does.

#include <stdio.h>
// Original (very incorrect)
#define SQR(x) x*x

// First correction (still incorrect)
#define SQR2(x) (x)*(x)

// second correction (still incorrect)
#define SQR3(x) ((x)*(x))

// correct
int square(int x){ return x*x; }

int main()
{
    int x1 = SQR(5);
    printf("5²=%d (correct so far)\n", x1);

    int x2 = SQR(1+1);
    printf("(1+1)²=%d (problem: operator precedence)\n", x2);
    
    int x3 = SQR2(1+1);
    printf("(1+1)²= %d (first correction)\n",x3);

    printf("sizeof (int) = %d\n", sizeof(int));
    int x4 = sizeof SQR2(1+1);
    printf("sizeof (1+1)²= %d (problem: operator precedence)\n", x4);
    
    int x5 = sizeof SQR3(1+1);
    printf("sizeof (1+1)²=%d (second correction)\n", x5);

    
    int i=2;
    int x6 = SQR3(i++);
    printf("(i++)²=%d (expected: 4, problem: undefined behavior)\n", x6);

    i=2;
    int x7 = SQR3(++i);
    printf("(++i)²=%d (expected: 9, problem: undefined behavior)\n", x7);

    i=2;
    int x8 = square(i++);
    printf("(i++)²=%d (expected: 4)\n",x8);

    i=2;
    int x9 = square(++i);
    printf("(++i)²=%d (expected: 9)\n",x9);
}

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