为什么按位“不是1”等于-2？ [英] Why does bitwise "not 1" equal -2?

问题描述

假设我们 1 ，基数2中的这个数字是：

Suppose we have 1 and this number in base 2 is:

00000000000000000000000000000001

现在我要翻转所有位以获得以下结果：

Now I want to flip all bits to get following result:

11111111111111111111111111111110

目前为止据我所知，解决方案是使用〜（按位NOT运算符）来翻转所有位，但结果为 ~1 是 -2 ：

As far as I know, the solution is to use the ~ (bitwise NOT operator) to flip all bits, but the result of ~1 is -2:

console.log(~1); //-2
console.log((~1).toString(2)); //-10 (binary representation)

为什么我会得到这个奇怪的结果？

Why do I get this strange result?

推荐答案

1 和 -2 ： 0 和 -1

1   二进制是 00000000000000000000000000000001

0   在二进制是 00000000000000000000000000000000

-1 二进制是 11111111111111111111111111111111

-2 二进制是 11111111111111111111111111111110

^{（binary是2的补码，在按位的情况下不是〜）}

1   in binary is 00000000000000000000000000000001
0   in binary is 00000000000000000000000000000000
-1 in binary is 11111111111111111111111111111111
-2 in binary is 11111111111111111111111111111110
^{("binary" being 2's complement, in the case of a bitwise not ~ )}

如你所见，这并不奇怪 ~1 等于 -2 ，因为〜0 等于 -1 。

As you can see, it's not very surprising ~1 equals -2, since ~0 equals -1.

正如 @Derek 解释，这些按位运算符将其操作数视为32位序列。另一方面， parseInt 则没有。这就是为什么你得到一些不同的结果。

As @Derek explained, These bitwise operators treat their operands as a sequence of 32 bits. parseInt, on the other hand, does not. That is why you get some different results.

这是一个更完整的演示：

Here's a more complete demo:

for (var i = 5; i >= -5; i--) {
  console.log('Decimal: ' + pad(i, 3, ' ') + '  |  Binary: ' + bin(i));
  if (i === 0)
    console.log('Decimal:  -0  |  Binary: ' + bin(-0)); // There is no `-0`
}

function pad(num, length, char) {
  var out = num.toString();
  while (out.length < length)
    out = char + out;
  return out
}

function bin(bin) {
  return pad((bin >>> 0).toString(2), 32, '0');
}

.as-console-wrapper { max-height: 100% !important; top: 0; }

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