如何在 Java 中创建 32 位掩码
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How to create 32 bit mask in Java
提问by uohzxela
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?
到目前为止，我有这段代码来创建 16 位掩码。但是我不知道如何使用这种方法创建 32 位掩码。有任何想法吗？
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...
编辑：我想创建 32 个 32 位掩码，每个掩码各自的位为 1，其余位为零。例如：掩码 1 的最左边的位为 1 而其余位为零，掩码 2 的第二个最左边的位为 1 而其余位为零。我不知道如何更简洁地解释，但我希望你们明白...
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
采纳答案by Stephen C
Here's how to create a 32 bit mask in Java.
这是在 Java 中创建 32 位掩码的方法。
int mask = 0x00010000; // for example.
And if you want to create a 32 bit mask with bit N set then
如果你想创建一个设置了位 N 的 32 位掩码，那么
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.
如果你想创建一个掩码数组，那么只需以明显的方式在循环中执行上述操作。
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 intis a 32 bit type in Java.
实际上，您假设的“16 位掩码”也是 32 位掩码，因为int是 Java 中的 32 位类型。
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?
和@Matt Ball 一样，我对您真正想做的事情感到困惑，以及您是否打算这样做是实现它的明智方法。当您可以用更少的代码即时创建掩码时，为什么需要一组掩码？
回答by Syd Kerckhove
I think you are doing well.
我认为你做得很好。
AMOUNT_OF_BITS = 32;
mask = new long[AMOUNT_OF_BITS];
for (long i = 0; i < AMOUNT_OF_BITS; i++)
{
mask[i] = Math.pow(2,i);
}
This should work, I think.
这应该有效，我想。
