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Ok, I think this should do the trick:

return((a ^ b) | (a ^ (*rp)) < 0) ? 1 : 0;

doesn't it?

Not yet. Consider a = -1, b = 1. Due to the OR, any a and b with different signs will return overflow.

Ok this is my dumbest time in life xD (I just directly didn't take into consideration that part because it never has overflow...)...

Most people don't appreciate how complicated C's mix of signed and unsigned is, especially when undefined behaviour of signed overflow is included.

Definitely.

Child, this is why I always add unit tests.