Why is this code resolved to true?

Question

int main() {
    int a = 1;
    int b = 0;

    if (a = b || ++a == 2)
        printf("T: a=%i, b=%i", a, b);
    else
        printf("F: a=%i, b=%i", a, b);

    return 0;
}

Let's take a look at this simple code snippet. Result is: T: a=1, b=0

Why? (note a=b uses assignment operand, not comparison)

What I understand here, is that zero is assigned to a, then a is incremented to 1. 1 is not equal to 2. So result should indeed be a=1, b=0. But why is this condition evaluated to true? Neither of (a=b) or (++a == 2) is true ... What did I miss?

Here is other short program that prints F as expected:

int main() {
    int a = 1;
    int b = 0;

    if (a = b) printf("T"); else printf("F");

    return 0;
}

Answer 1

You have confused yourself with misleading spacing.

if (a = b || ++a == 2)

is the same as:

if (a = (b || ((++a) == 2)))

This actually has undefined behavior. Although there is a sequence point between the evaluation of b and the evaluation of ((++a) == 2), there is no sequence point between the implied assignment to a and the other write to a due to the explicit = assignment.

Answer 2

Actually, assignment has the lowest operator precedence so your if statement is equivalent to:

if ( a = ( b || ( ++a == 2 ) ) )

So you're assigning a to 1 but also incrementing it in the same expression. I think that leads to undefined behavior, but the end result is that a is 1 in your compiler.