Is this expression statement in a C programming exam question undefined behavior?

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In a C programming exam question, I found this:

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

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?

undefined-behavior c
2 Answers

Is this expression statement in a C programming exam question undefined behavior?

Is this expression UB?

No.

Any mistake?

No.

I wanted to write some text here, but you have already written the explanation... Nothing for me to do. Notes: There is also a sequence point after ?. Even if y-- would be evaluated, because && is a sequence point, it still would be fine.

user_90727530 
#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.

user_113095000
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