Java：将带符号的Int_32（十六进制）转换为Int

Question

I've got a signed int_32 value (FFFFFFFC) what's representing -4

If I try to convert it by ...

long x = Long.parseLong("FFFFFFFC", 32);
System.out.println("X: " + x);

... I'll get "X: 532021755372" instead of -4. How could I convert it as a signed value?

Answer 1

You get an incorrect result because 32 represents the numeric base , not the number of bits in the result. You are parsing the string as a base-32 number (ie a number in a numbering system that uses digits 0..9 and letters A..V, not 0..9 and A..F.

To parse the number correctly, use Long.parseLong("FFFFFFFC", 16); , and cast the result to int , which is a 32-bit number:

int x = (int)Long.parseLong("FFFFFFFC", 16);
System.out.println("X: " + x); // Prints X: -4

Demo.

Answer 2

if FFFFFFFC represents -4 it is int, not long, parse it like this

int x = Integer.parseUnsignedInt("FFFFFFFC", 16);

and you will get -4

Answer 3

Try this

int x= new BigInteger("FFFFFFFC", 16).intValue();
System.out.println("X: " + x);