Java Integer index to bit position

Question

I'm trying to make a dynamic method to check a number through binary anding. I've tried a variety of different approaches but just can't wrap my head around it.

what I need is a method that converts an integer index position and returns a boolean.

private boolean bitwiseAnding(int val, int indx){
    //TODO Convert indx to a bitmask here
    return (val & bitmask)==1;
}

for example:

1 = 0x01
2 = 0x02
3 = 0x04
4 = 0x08
and so on

is this not the same as rotate 1 n places to the left? – ΦXocę 웃 Пepeúpa ツ Oct 25 '16 at 12:27 — ΦXocę 웃 Пepeúpa ツ, Oct 25 '16 at 12:27

Krzysztof Atłasik · Accepted Answer · 2016-10-25T12:31:22.353

5

You have to use bit shifting operator:

private boolean bitwiseAnding(int val, int indx){
    int mask = 1 << (indx-1);
    return (val & mask) != 0;
}

edited Oct 25 '16 at 12:31

answered Oct 25 '16 at 12:29

Krzysztof Atłasik

21,985
6
54
76

1

This fails for `indx == 31`. – Marko Topolnik Oct 25 '16 at 12:30

score 5 · Answer 2 · answered Oct 25 '16 at 12:30

5

Just use bit shifting:

int bitmask = 1 << (indx-1);

But note that you'll only get true for ==1 if indx == 1. You might mean:

(val & bitmask) != 0

answered Oct 25 '16 at 12:30

Andy Turner

137,514
11
162
243

Java Integer index to bit position

2 Answers2