28 ms·
Suppose we want to store 2^(48-n) 2^n byte objects in your hypothetical <42 bits of address space. Say, 2^47 two byte objects. (Alternatively, more than 2^(48-n
by AaronFriel 8y ago
Suppose we want to store 2^(48-n) 2^n byte objects in your hypothetical <42 bits of address space. Say, 2^47 two byte objects. (Alternatively, more than 2^(48-n) 2^n byte objects in 42 bits of address space.)
How do I compute the hash of every object, since by definition I can't address every byte?
- chrisseaton 8y agoYou can still read any byte you want by decompressing the pointer and using it as normal. And why would you want to read raw bytes from an object to hash it? What requires this in Java?