LargeInteger等于BigInteger的testBit吗? [英] LargeInteger's equivalent of BigInteger's testBit?

问题描述

LargeInteger 是否具有与 BigInteger的testBit ?

如果没有，如何在LargeInteger上执行testBit?

If not, how can testBit be performed on a LargeInteger?

我还没有复制((this & (1<<n)) != 0)的必要技能.

I don't yet have the necessary skills to reproduce ((this & (1<<n)) != 0).

我尝试通过仅复制& ;;粘贴文档中引用的上述代码:

I tried making a method by just copying & pasting the above code referenced from the docs:

static boolean testBit(int n){
    return ((this & (1<<n)) != 0);
}

但是，编译器报告:

error: non-static variable this cannot be referenced from a static context
    return ((this & (1<<n)) != 0);
             ^
error: bad operand types for binary operator '&'
    return ((this & (1<<n)) != 0);
                  ^

推荐答案

在给定API的情况下，这是我能提供的最好的解决方案:

This is the best I can come up with given the API:

static boolean testBit(LargeInteger i, int n) {
    return i.shiftRight(n).isOdd();
}

n是要测试的位的位置.

n is the position of the bit to be tested.

我假设您将此方法放在某个实用程序类中.

I assume you put this method in some utility class.

通常，您将执行num & (1 << pos)提取pos位置的位:

Normally, you would do num & (1 << pos) to extract the bit at pos position:

???????x?????
0000000100000
-------------
0000000x00000

如果整个事物为0，则x为0；如果整个事物为0，则x为0.否则，x为1.

If the whole thing is 0 then x is 0; otherwise, x is 1.

在上述方法中，我执行num >> pos:

In the method above, I do num >> pos:

???????x?????
-------------
????????????x

我们知道二进制数的最低有效位为1时为奇数，而最低有效位为0时则为偶数.

We know that a binary number is odd when its least significant bit is 1, and it is even when its least significant bit is 0.

因此，如果右移后的数字为奇数，则我们知道该位为1；即使是偶数，我们也知道该位为0.

So if the number after right-shifting is odd, we know the bit is 1; if even, we know the bit is 0.

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