Why is a pointer to a pointer to a function 8 byte?

Question

I know a pointer to a function is 8 byte because of virtualization but why a pointer to a pointer to a function is 8 byte?

typedef void(*fun())();
sizeof(fun*); // returns 8 byte

Answer 1

If you have a 64-bit system with 8-bit bytes (and it sounds like you do), probably all pointers will be 8 bytes in size. Virtualization doesn't have anything to do with it.

Answer 2

It's because they're both pointers to memory addresses regardless of what kind of data they point to, and you're running on a 64-bit system

If memory addresses were say... 4 bytes, than it would be impossible to have more than 4GB of ram on your computer. there just wouldn't be enough different pointer values.