7 ms·
Sorry, I'm not sure how this is a problem. On segmented architectures size_t is smaller than uintptr_t, but that just means there needs to be an allocation limi
by TheCycoONE 5y ago
Sorry, I'm not sure how this is a problem. On segmented architectures size_t is smaller than uintptr_t, but that just means there needs to be an allocation limit < usize::max_value.
It would have cause more bugs if they defined it the other way and people used usize to store addresses.