Ternary Operators and Macros #define

Question

I am confused on the ternary operator outputs if it is used in a macro. Why is the output 3 and 0?

#define test(x,y)(! y ? 0 : x/y)

int main(void){
printf("%d",test(2+2,2));
printf("%d",test(4,-1-1));
return(0);
}

Answer 1

When you are confused about the usage of a macro, copy-paste it where it is being used and evaluate from there:

int main(void){
    printf("%d", ! 2 ? 0 : 2+2 / 2);
    printf("%d", ! -1 - 1 ? 0 : 4 / -1 - 1);
    return(0);
}

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The second argument to the first printf call goes: if ! 2 is true, return 0, otherwise return the expression (2+2) / 2. Since ! 2 is false, the expression 2+2 / 2 is returned, and its result is clearly 3.

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The second argument to the second printf call goes: if ! -1 - 1 is true, return 0, otherwise return the expression 4 / -1 - 1.

The expression ! -1 -1 is actually interpreted as !(-1) - 1, since the ! (logical NOT) operator has precedence over the - (subtraction) operator.

!(-1) becomes 0 (-1 is a truthy value, which, when negated, becomes a falsy one, i.e. 0).

Then, 0 - 1 becomes -1, which is a truthy value, therefore the truthy value of the ternary operator is returned, i.e. 0.

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