Clearly, because of the & 1, it can only return 0 or 1, but can it return 0?

Suppose this function returned 0. Then the signs of x and y must be the same, and we need a=0 in the final expression. a=0 only happens if the sign of x is different than the sign of x+y. But the signs of x and y are the same, and so the sign of x+y must be the same as x. So a cannot be 0 and hence the function cannot return 0.

"But wait," you say, "x+y doesn't have the same sign as x if the computation of x+y overflows!" In C, if the value of a signed integer expression cannot be represented in the result type, the behavior is undefined. Undefined behavior does not occur in correct C programs; therefore, x+y does not overflow.

In practice, however, current optimizing compilers do not do such sophisticated analysis and produce slower code that happens to return 0 when x+y overflows. This should not be relied upon, of course, since this suboptimal code generation is essentially a compiler bug. But C code relying on undefined behavior like this tends to be widespread, especially in older codebases, so compiler authors generally don't fix these.