How to create 32 bit mask in Java

Question

So far I have this code to create 16 bit mask. However I do not know how to create a 32 bit mask using this approach. Any ideas?

edit: I want to create 32 masks of 32 bits, each with with its respective bit being 1 and the rest of the bits being zero. For example: mask 1 has the leftmost bit being 1 while the rest of the bits are zero, mask 2 has the 2nd leftmost bit being 1 while the rest of the bits are zero. I dunno how to explain more succinctly but I hope you guys get the idea...

mask = new int[16];
mask[0] = 0x8000;
mask[1] = 0x4000;
mask[2] = 0x2000;
mask[3] = 0x1000;

mask[4] = 0x0800;
mask[5] = 0x0400;
mask[6] = 0x0200;
mask[7] = 0x0100;

mask[8] = 0x0080;
mask[9] = 0x0040;
mask[10] = 0x0020;
mask[11] = 0x0010;

mask[12] = 0x0008;
mask[13] = 0x0004;
mask[14] = 0x0002;
mask[15] = 0x0001

Solution

Here's how to create a 32 bit mask in Java.

int mask = 0x00010000;  // for example.

And if you want to create a 32 bit mask with bit N set then

int mask = 1 << N;  // Assumes the rightmost bit is numbered zero ...

And if you want to create an array of masks, then just do the above in a loop, in the obvious fashion.

int[] masks = new int[32];
for (int n = 0; n < 32; n++) {
    masks[n] = 1 << n;
}

---

In reality, your supposed "16 bit masks" are also 32 bit masks, because int is a 32 bit type in Java.

And like @Matt Ball, I'm puzzled as to what you are really trying to do, and whether you are going about it is a sensible way to achieve it. Why do you need an array of masks when you can create a mask on the fly with less code?

Comments

[uohzxela]

Sorry for the unclear question. I edited it. Hope it makes sense now.

[Dolda2000]

"Assumes the leftmost bit is numbered zero ..." -- It actually assumes the rightmost bit is numbered zero. ;)

[uohzxela]

Many thanks. Your answer clears some air and steered me in the right direction.

[Aaron Dancygier]

This won't work Math.pow(x,y) returns double. You will have to cast to long. mask[i] = (long) Math.pow(2,i);