I have a string as "5F2A" as Hex. I would like to convert it as int 0x5F2A.
String str = "5F2A";
int number = someOperation(str);
And the number should be (with 0x)
0x5F2A
Is it possible?
To rephrase and share what I learnt today
Map<Integer, String> map = new HashMap<>();
map.put(0x5F2A, "somevalue");
System.out.println(map.get(24362));
System.out.println(map.get(0b0101111100101010));
Would give the value somevalue for both.
No transformation required:
System.out.println("0x" + str);
And to turn an arbitrary int into HEX representation:
Integer.toHexString(intNumber);
That should be all you need to get going!
Integer.parseInt(String, int).int n = 0x5F2A;
0x5F2A is only a represenation of number 24362. And this is only a presentation for binary 101111100101010.int i = 0x5F2A not really means nothing because in memory, all is in binary, it's only when you print that it matters
String str = "5F2A";
int number = Integer.parseInt(str, 16); //alows to store an int, binary 0101111100101010
System.out.println(number); //24362 (decimal by default)
System.out.println(Integer.toHexString(number)); //5f2a (hexa possible too)
By default, it prints in (binary into) decimal format, but you can print in hexa format, but int i = 0x5F2A means at 100% the same as int i = 24362
System.out.println(number);will always output in decimal, not hexadecimal. You can useSystem.out.format.int i = 0x5F2A;int number = Integer.parseInt(str, 16);is transforming aStringto anintwith base 16(hex).0x5F2A.