如何检查是否一个数是2的幂

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问题描述

今天我需要一种简单算法检查是否一个数是2的幂。

Today I needed a simple algorithm for checking if a number is a power of 2.

该算法需要：

简单
校正任何 ULONG 值。

我想出了这个简单的算法：

I came up with this simple algorithm:

private bool IsPowerOfTwo(ulong number)
{
    if (number == 0)
        return false;

    for (ulong power = 1; power > 0; power = power << 1)
    {
        // This for loop used shifting for powers of 2, meaning
        // that the value will become 0 after the last shift
        // (from binary 1000...0000 to 0000...0000) then, the 'for'
        // loop will break out.

        if (power == number)
            return true;
        if (power > number)
            return false;
    }
    return false;
}

不过转念一想，怎么样，如果检查日志 X 是一个确切的轮数？但是，当我检查了2 ^ 63 + 1，将Math.log 返回正好63由于四舍五入的原因。所以，我检查，如果2电源63等于原来的号码 - 这是因为计算是做双和没有确切的数字：

But then I thought, how about checking if log x is an exactly round number? But when I checked for 2^63+1, Math.Log returned exactly 63 because of rounding. So I checked if 2 to the power 63 is equal to the original number - and it is, because the calculation is done in doubles and not in exact numbers:

private bool IsPowerOfTwo_2(ulong number)
{
    double log = Math.Log(number, 2);
    double pow = Math.Pow(2, Math.Round(log));
    return pow == number;
}

这回真对于给定的错误的值： 9223372036854775809

This returned true for the given wrong value: 9223372036854775809.

有没有更好的算法？

说明

首先按位二进制和放大器;从MSDN定义操作符：

Explanation

First and foremost the bitwise binary & operator from MSDN definition:

二元放大器;运营商pdefined为整型和布尔$ P $。对于整型，＆安培;计算其操作数的逻辑位与。对于bool操作数，及放大器;计算逻辑与的操作数;那是，则结果为真，当且仅当两个操作数都为真。

现在，让我们来看看这一切是如何发挥出来：

Now let's take a look at how this all plays out:

该函数返回布尔型（真/假）和接受类型无符号长的一个传入确定参数（x，在这种情况下）。让我们为简单起见，假设有人通过了价值4，并呼吁像这样的功能：

The function returns boolean (true / false) and accepts one incoming parameter of type unsigned long (x, in this case). Let us for the sake of simplicity assume that someone has passed the value 4 and called the function like so:

bool b = IsPowerOfTwo(4)

现在我们替换x的每次出现与4：

Now we replace each occurrence of x with 4:

return (4 != 0) && ((4 & (4-1)) == 0);

嗯，我们已经知道，4！= 0 evals为true，到目前为止，一切顺利。但怎么样：

Well we already know that 4 != 0 evals to true, so far so good. But what about:

((4 & (4-1)) == 0)

这相当于本课程的：

This translates to this of course:

((4 & 3) == 0)

但究竟什么是 4安培; 3

的4二进制重新presentation是100和3的二进制重新presentation是011（记得＆安培;将这些数字的二进制重新presentation因此，我们有：

The binary representation of 4 is 100 and the binary representation of 3 is 011 (remember the & takes the binary representation of these numbers. So we have:

100 = 4
011 = 3

想象一下，这些值被堆积起来很像小学加法。该＆安培; 运营商说，如果两个值等于1，则结果为1，否则为0，所以 1安培; 1 = 1 ， 1安培; 0 = 0 ， 0安培; 0 = 0 和 0安培; 1 = 0 。所以我们做数学题：

Imagine these values being stacked up much like elementary addition. The & operator says that if both values are equal to 1 then the result is 1, otherwise it is 0. So 1 & 1 = 1, 1 & 0 = 0, 0 & 0 = 0, and 0 & 1 = 0. So we do the math:

100
011
----
000

其结果仅仅是0，所以，我们回去看看退货语句现在翻译为：

The result is simply 0. So we go back and look at what our return statement now translates to:

return (4 != 0) && ((4 & 3) == 0);

现在翻译为：

Which translates now to:

return true && (0 == 0);

return true && true;

我们都知道，真放;＆安培;真正的简直就是真，这说明我们的例子中，4是2的力量。

We all know that true && true is simply true, and this shows that for our example, 4 is a power of 2.

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