3 ms·
> uint32_t gets converted to a 64-bit signed value before any arithmetic operator is applied. Only if int is > 32 bits on your platform, which is quite rare th
by stephencanon 11y ago
> uint32_t gets converted to a 64-bit signed value before any arithmetic operator is applied.
Only if int is > 32 bits on your platform, which is quite rare these days.