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It's not true that (unsigned)INT_MAX + 1 is never 0 - it's allowed for UINT_MAX == INT_MAX (leaving the sign bit unused in the unsigned representation). I've n
by caf 14y ago
It's not true that (unsigned)INT_MAX + 1 is never 0 - it's allowed for UINT_MAX == INT_MAX (leaving the sign bit unused in the unsigned representation). I've never seen such an implementation, though.
- jacquesgt 14y agoYep, you're right.