3 ms·
lol. You can't always subtract two size_t's. (size_t)3 - (size_t)4. And (ptrdiff_t)((size_t)3-(size_t)4) actually would be undefined behavior in C++20!
by rerdavies 2y ago
lol. You can't always subtract two size_t's. (size_t)3 - (size_t)4. And (ptrdiff_t)((size_t)3-(size_t)4) actually would be undefined behavior in C++20!
- Joker_vD 2y agoThe "(size_t)3 - (size_t)4" is defined to be the same as "-(size_t)1", which is defined, and the same as "(size_t)1 + ~(size_t)1". Also, "(ptrdiff_t)((size_t)3-(size_t)4)" is defined both before and after C++20: If the destination type is signed, the value does not change if the source integer can be represented in the destination type. Otherwise the result is [implementation-defined (until C++20)] [the unique value of the destination type equal to the source value modulo 2^n where n is the number of bits used to represent the destination type (since C++20)] (note that this is different from signed integer arithmetic overflow, which is undefined).
- rerdavies 2y agoHowever, the value of (ptrdiff_t)((size_t)3-(size_t)4) is positive if ptrdiff_t has at least 17 bits, and size_t only has 16 bits, and negative whenever ptrdiff_t and size_t have the same number of bits.