Javascript波浪号(~)将给定值强制转换为1的补位——所有位都是反向的。 这就是波浪的作用。这不是固执己见。它既不加也不减任何量。

0 -> 1
1 -> 0
...in every bit position [0...integer nbr of bits - 1]

On standard desktop processors using high-level languages like JavaScript, BASE10 signed arithmetic is the most common, but keep in mind, it's not the only kind. Bits at the CPU level are subject to interpretation based on a number of factors. At the 'code' level, in this case JavaScript, they are interpreted as a 32-bit signed integer by definition (let's leave floats out of this). Think of it as quantum, those 32-bits represent many possible values all at once. It depends entirely on the converting lens you view them through.

JavaScript Tilde operation (1's complement)

BASE2 lens
~0001 -> 1110  - end result of ~ bitwise operation

BASE10 Signed lens (typical JS implementation)
~1  -> -2 

BASE10 Unsigned lens 
~1  -> 14 

以上所有观点同时都是正确的。

Javascript波浪号(~)将给定值强制转换为1的补位——所有位都是反向的。 这就是波浪的作用。这不是固执己见。它既不加也不减任何量。

0 -> 1
1 -> 0
...in every bit position [0...integer nbr of bits - 1]

On standard desktop processors using high-level languages like JavaScript, BASE10 signed arithmetic is the most common, but keep in mind, it's not the only kind. Bits at the CPU level are subject to interpretation based on a number of factors. At the 'code' level, in this case JavaScript, they are interpreted as a 32-bit signed integer by definition (let's leave floats out of this). Think of it as quantum, those 32-bits represent many possible values all at once. It depends entirely on the converting lens you view them through.

JavaScript Tilde operation (1's complement)

BASE2 lens
~0001 -> 1110  - end result of ~ bitwise operation

BASE10 Signed lens (typical JS implementation)
~1  -> -2 

BASE10 Unsigned lens 
~1  -> 14 

以上所有观点同时都是正确的。

很简单:

Before starting please remember that 
 1  Positive numbers are represented directly into the memory.
 2. Whereas, negative numbers are stored in the form of 2's compliment.
 3. If MSB(Most Significant bit) is 1 then the number is negative otherwise number is 
    positive.

你会发现~2:

Step:1 Represent 2 in a binary format 
       We will get, 0000 0010

Step:2 Now we have to find ~2(means 1's compliment of 2)
                  1's compliment       
       0000 0010 =================> 1111 1101 

       So, ~2 === 1111 1101, Here MSB(Most significant Bit) is 1(means negative value). So, 
       In memory it will be represented as 2's compliment(To find 2's compliment first we 
       have to find 1's compliment and then add 1 to it.)

Step3:  Finding 2's compliment of ~2 i.e 1111 1101

                   1's compliment                   Adding 1 to it
        1111 1101 =====================> 0000 0010 =================> 0000 0010
                                                                      +       1
                                                                      ---------
                                                                      0000 0011 
        So, 2's compliment of 1111 1101, is 0000 0011 

Step4:  Converting back to decimal format.
                   binary format
        0000 0011 ==============> 3
        
       In step2: we have seen that the number is negative number so the final answer would  
       be -3
                                    
                                So, ~2 === -3

首先，我们必须把给定的数字分成它的二进制数，然后把它颠倒过来，把最后一个二进制数相加。执行完后，我们必须给我们正在寻找补数的前一位数字赋相反的符号 ~ 2 = 3 解释: 2的二进制形式是00000010变成11111101，这是1的补码，然后补码为00000010+1=00000011，这是3的二进制形式，带-符号，即-3

int = 4; System.out.println (~); 结果是:-5

Java中任意整数的“~”表示1对no的补。 例如，我取~4，这意味着用二进制表示0100。 首先, 整数长度为4字节，i。e4 *8(8位1字节)=32。 在系统内存中，4表示为 0000 0000 0000 0000 0000 0000 0000 0100 现在~操作符将对上面的二进制no执行1的补

i.e 1111 1111 1111 1111 1111 1111 1111 1011->1's complement the most significant bit represents sign of the no(either - or +) if it is 1 then sign is '-' if it is 0 then sign is '+' as per this our result is a negative number, in java the negative numbers are stored in 2's complement form, the acquired result we have to convert into 2's complement( first perform 1's complement and just add 1 to 1's complement). all the one will become zeros,except most significant bit 1(which is our sign representation of the number,that means for remaining 31 bits 1111 1111 1111 1111 1111 1111 1111 1011 (acquired result of ~ operator) 1000 0000 0000 0000 0000 0000 0000 0100 (1's complement)

1(2的补数)

1000000 0000 0000 0000 0000 0000 0000 0101 现在结果是-5 查看视频<[java中的位运算符]https://youtu.be/w4pJ4cGWe9Y的链接

我知道这个问题的答案很久以前就贴出来了，但我想分享我的答案。

要找到一个数的一补，首先要找到它的二进制等价物。这里，十进制数字2用二进制形式表示为0000 0010。现在通过将其二进制表示的所有数字逆(将所有1都翻转为0，将所有0都翻转为1)来求其1的补数，这将得到:

0000 0010 → 1111 1101

这是十进制数2的1补。由于二进制数的第一个位，即符号位为1，这意味着它存储的数字的符号为负。(这里所指的数字不是2，而是2的1的补数)。

现在，由于数字存储为2的补数(取1的补数加1)，所以要将这个二进制数1111 1101显示为十进制，首先我们需要找到它的2的补数，即:

1111 1101 → 0000 0010 + 1 → 0000 0011

这是2的补。二进制数0000 0011的十进制表示是3。并且，因为符号位是1，所以结果是-3。

提示:如果你仔细阅读这个过程，你会发现1的补码操作符的结果实际上是，数字(操作数-，这个操作符被应用)加1，带一个负号。你也可以用其他数字试试。