In a C programming exam question, useless and unreasonable for my standards, I found this:

int a, b=0, x=4, y=5;
a=((a=x%y?b 1:y--)&&(x-=y))||(y-=6);

I know that the correct behavior would be to shoot the programmer who wrote this, but cope with me. Is this expression UB?

I would say no, because of the sequence point (SP) between logic operators and between expressions of ?:. So, to my understanding, the horrible hack would be evaluated correctly as:

      x%y                             this gives 4 (exp is true)
    a=    b 1                         this assigns 1 to a and we have a SP (exp is true)
              y--                     this is not evaluated
                     x-=y             this assigns -1 to x and we have a SP (exp is true) 
                              y-=6    this is not evaluated
a=                                    this assigns the result of the || operator to a (1)

Finally a=1, x=-1, nothing else is changed. Any mistake?

CodePudding user response：

#include <stdio.h>

int main() {
    int a, b=0, x=4, y=5;
    a=((a=x%y?b 1:y--)&&(x-=y))||(y-=6);
    printf("a=%i\nb=%i\nx=%i\ny=%i\n", a,b,x,y);
    return 0;
}

OUTPUT:

a=1
b=0
x=-1
y=5

So I think you are right about everything but I do not think it counts as undefined behavior.

CodePudding user response：

I think, that expression has a completely defined behaviour.

In C, boolean expressions are evaluated from left to right.

And in Boolean algebra && have higher precedence than ||.

So (a=x%y?b 1:y--)&&(x-=y)) will evaluate first and within it (a=x%y?b 1:y--) gets evaluated first.