How does C read an equality operator in the initialization of the counter of a "for" loop?

int main() {
    int x = 0;
    int i;
    for (i == 0; i < 3; i++) {
        x = x * 2;
    }
}

This code runs without errors and I'm not sure why. When I initialize the loop counter, I put the equality symbol instead of just the equal sign, yet it still treats i as 0. Why does C not recognize that I made a mistake here?

Your code has undefined-behavior.

It might seem to work well but you cannot rely on that.

i is never initialized in your code and therefore has an indetermined value.

The initialization clause of the loop check i's indetermined value against 0 (and the result of the comparison is discarded anyway).

Then the uninitialized i is read for checking if the loop should continue to the next iteration (with i < 3) and finally incremented with i++ (if the former condition will be evaluated to true).

Both of these will invoke undefined-behavior, and therefore no outcome is guaranteed.

In any case to answer your question in the title:

How does C read an equality operator in the initialization of the counter of a "for" loop?

Using the comparison operator there would be discarded anyway, since the outcome of the comparison has no effect.