4 ms·
2^(bitsize) == 0. Edit: Yes, 1 << bitsize is undefined. But unsigned integers actually do have well-defined semantics on overflow, and multiplying by 2 enoug
by wnoise 15y ago
2^(bitsize) == 0.
Edit: Yes, 1 << bitsize is undefined. But unsigned integers actually do have well-defined semantics on overflow, and multiplying by 2 enough times really does produce zero.