Java convert 16 bits signed to 4 digits hex

Question

When dealing with signed int conversion to hex, only Integer.toString(value, 16) is useful

see a post on subject but I need to format with 4 hex digits (leading zeros for positive numbers and not 32bits/8chars for negatif numbers),

But I didn't fiund the equivalent in Java (Android).

4 hex digits would be the type short (16 bits) not Integer (32) — Tony Hopkinson, May 18 '14 at 01:40
found it : int StepRef =-2; String SS = String.format("%08X",StepRef); String SS4 = SS.substring(SS.length() - 4); — Lotfi, May 18 '14 at 01:48
Are you sure that's what you want to do. It's only converting "half" the number". — Tony Hopkinson, May 18 '14 at 01:57
yes, I need only 16 bits wide ! it works like IntToHex( value,4) — Lotfi, May 18 '14 at 03:17
Yeah but if for some reason in your code you "overflow" e.g 32766 = "0000", finding out where and why is likely to be painful. — Tony Hopkinson, May 18 '14 at 03:22
IntToHex() is not C++. It is just a functions that comes with C++ Builder. — greenapps, May 18 '14 at 09:06
my int is 16 bits, coming from dspic 16 bits, it can't overflow — Lotfi, May 18 '14 at 19:09

score 0 · Answer 1 · answered May 18 '14 at 19:08

0

Found it :

int StepRef =-2;
String SS = String.format("%08X",StepRef);
String SS4 = SS.substring(SS.length() - 4);

answered May 18 '14 at 19:08

Lotfi

1 Answers1