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Do signed types not also "break trivial math" like that, just at a different boundary? Genuine question. (The 0 boundary is obviously going to be more commonly

by mattdw 11y ago

Do signed types not also "break trivial math" like that, just at a different boundary? Genuine question. (The 0 boundary is obviously going to be more commonly hit than the 2^32 boundary, but nonetheless.)

mohawk 11y ago
Signed types have the same "wraparound" problem, i think the OP meant that they don't have this problem at the zero boundary.
- mikeash 11y ago
  Signed types have a worse problem. They typically wrap around, but the behavior is undefined. That means the compiler can assume it never happens and optimize your code accordingly, which can lead to all sorts of entertaining misbehavior.
- mohawk 11y ago
  Sorry meant two's complement
zvrba 11y ago
Yes, except that you're far more often operating at the 0 boundary instead of at the INT_MIN / INT_MAX boundaries. Also note that C only half-heartedly supports objects larger than SIZE_MAX/2: relevant quote from http://en.cppreference.com/w/cpp/types/ptrdiff_t http://en.cppreference.com/w/cpp/types/ptrdiff_t "If an array is so large (greater than PTRDIFF_MAX elements, but less than SIZE_MAX bytes), that the difference between two pointers may not be representable as std::ptrdiff_t, the result of subtracting two such pointers is undefined. "