4 ms·

What would you do if your array is so large that it requires an unsigned int64 index?

• OptionOfT 11mo ago
  The current AMD64 specification only uses 48-bits of pointer space, coming from 40-bits. So we still have 16 bits remaining. I'm sure we can use 1 for a sign.
• adrian_b 11mo ago
  In C/C++ there are no true unsigned integers (true unsigned integers do overflow, generating errors in such cases or they generate carry/borrow on certain operations). The so-called unsigned integers of C/C++ are in fact modular integers, i.e. where the arithmetic operations wrap around and where you can interpret any 64-bit number greater than 2^63-1 as you please, as either a positive integer or as a negative integer. For instance you can interpret 2^63 as either -2^63 or as +2^63. So using 64-bit "unsigned" integers for indices does not create any problems if you are careful how you interpret them. However, as another poster has already said, in all popular ISAs the addressable space is actually smaller than 2^64 and in x86-64 the addresses are interpreted as signed integers, not as unsigned integers, so your problem can never appear. Some operating systems use this interpretation of the addresses as signed integers in order to reserve the negative addresses for themselves and use only positive addresses for the non-privileged programs. The reason why the addressable space is smaller than afforded by 64 bits is that covering the complete space with page translation tables would require too many table levels. x86-64 has increased the number of page table levels to 4, in order to enable a 48-bit address space, while some recent Intel/AMD CPUs have increased the number of page table levels to 5, in order to increase the virtual address size to 57 bits.