混淆PHP按位NOT行为

问题描述：

在PHP中，如果我运行以下简单程序

In PHP, if I run the following simple program

$number = 9;
var_dump( ~ $number );

我的输出是

int(-10)

这让我感到困惑.我认为 ~是按位的NOT运算符.所以我期待着类似的事情.

This is confusing to me. I thought ~ was the bitwise NOT operator. So I was expecting something like.

if binary 9 is     00000000000000000000000000001001 
then Bitwise NOT 9 11111111111111111111111111110110

含义~9应该以类似于4,294,967,286的一些荒谬的大整数形式出现.

Meaning ~9 should come out as some ludicrously large integer like 4,294,967,286.

什么是优先级，强制类型，或者我在这里还缺少其他什么?

What subtly of precedence, type coercion, or something else am I missing here?

答

您的输出默认为带符号的int-将其包装在decbin中以获取二进制表示形式.

Your output is defaulting to a signed int - wrap it in decbin to get a binary representation.

考虑:

$number = 9;
var_dump(  bindec(decbin(~ $number)) );

With two's compliment, the MSB of a signed binary number becomes 0-MSB, but every other bit retains its respective positive values.

因此，为了论证(一个8位示例)，

So for argument's sake (an 8-bit example),

Binary 9: 0000 1001
Inverse:  1111 0110

结果为(-128) + 64 + 32 + 16 + 4 + 2 = -10，因此PHP可以正确计算，只需对MSB应用2的补码即可.

This results in (-128) + 64 + 32 + 16 + 4 + 2 = -10, so PHP is calculating correctly, its just applying two's compliment to the MSB.