3 ms·

You also have to account for any shared objects that might get mapped in, the executable data, the stack, and the heap. AFAIK, usually only the low 32 address b

by gary_0 5y ago

You also have to account for any shared objects that might get mapped in, the executable data, the stack, and the heap. AFAIK, usually only the low 32 address bits are used for all that, leaving us with 46 bits.

moonchild 5y ago
Going from 47 bits to 46 bits would mean you've spent half your address space on something. 2^32 is much less than half of 2^47; you still have the _vast majority_ of a 47-bit address space.
- gary_0 5y ago
  Ah, right. I can't math.