Why f(0) + f(11) == f(0) in C language by conditional define f(0)

Question

#include<stdio.h>
#define f(x) (x<10) ? 1 : 3

int main(){
    int i = f(1) + f(11);
    printf("%d", i);
    return 0;
}

Why in this case, the program output "1" but not "4".
How exactly do these two different uses of the defined macro work?

Answer 1

Look at what the compiler sees when pre-processing is done, with some () added for clarity:

#include<stdio.h>

int main(){
    int i = (1<10) ? 1 : (3 + (11<10) ? 1 : 3);
    printf("%d", i);
    return 0;
}

The reason for the () where I added them can be read here:
https://en.cppreference.com/w/c/language/operator_precedence

It shows that ?: has low priority and the + is pushing in to influence the result being evaluated for the first : part.

Now it should be clear.

The solution is to generously use () within macro definitions.
Comments and other answers have already provided that, I repeat for completeness:

#define f(x) (((x)<10) ? 1 : 3)

I seem to have even more () than other proposals.
Mine also protects against funny things like f(12||12) or f(0&&0) behaving weirdly.