Java： ~ 是什么意思

Java: What does ~ mean

提问by Martijn Courteaux

In this Java source code I have this line:

在这个 Java 源代码中，我有这一行：

if ((modifiers & ~KeyEvent.SHIFT_MASK) != 0) ....

What does the tilde ~mean?

波浪号~是什么意思？

采纳答案by Richard

The Tilde (~) performs a bitwise complement of a numerical value in Java.

波浪号 ( ~) 在 Java 中执行数值的按位补码。

See: Bitwise complement (~): inverts ones and zeroes in a number

请参阅：按位补码 ( ~)：将数字中的 1 和 0 取反

回答by Pascal MARTIN

It is the Unary ~ Bitwise complementoperator (quoting):

它是一元 ~ 按位补码运算符（引用）：

• only used with integer values
• inverts the bits ie a 0-bit becomes 1-bit and vice versa
• in all cases ~x equals (-x)-1

• 仅用于整数值
• 反转位，即 0 位变为 1 位，反之亦然
• 在所有情况下 ~x 等于 (-x)-1

See also this page on Bitwise operators on wikipedia, which states :

另请参阅维基百科上有关按位运算符的此页面，其中指出：

The bitwise NOT, or complement, is a unary operation that performs logical negation on each bit, forming the ones' complement of the given binary value. Digits which were 0 become 1, and vice versa.
For example:

按位非或补码是一种一元运算，它对每一位执行逻辑否定，形成给定二进制值的补码。0 的数字变为 1，反之亦然。
例如：

NOT 0111  (decimal 7)
  = 1000  (decimal 8)

In many programming languages (including those in the C family), the bitwise NOT operator is "~" (tilde).

在许多编程语言 （包括 C 系列中的那些）中，按位非运算符是“ ~”（波浪号）。

回答by Alberto Zaccagni

From the official docs http://java.sun.com/docs/books/tutorial/java/nutsandbolts/op3.html:

从官方文档http://java.sun.com/docs/books/tutorial/java/nutsandbolts/op3.html：

The unary bitwise complement operator "~" inverts a bit pattern; it can be applied to any of the integral types, making every "0" a "1" and every "1" a "0". For example, a byte contains 8 bits; applying this operator to a value whose bit pattern is "00000000" would change its pattern to "11111111".

一元按位补码运算符“~”反转位模式；它可以应用于任何整数类型，使每个“0”成为“1”，每个“1”成为“0”。例如，一个字节包含 8 位；将此运算符应用于位模式为“00000000”的值会将其模式更改为“11111111”。

回答by VolkerK

As said before ~is the unary bitwise NOT operator.
Your example tests whether modifierscontains bits other than those defined in KeyEvent.SHIFT_MASK.

如前所述~是一元按位非运算符。
您的示例测试是否modifiers包含中定义的位以外的位KeyEvent.SHIFT_MASK。

• ~KeyEvent.SHIFT_MASK-> all bits except those in KeyEvent.SHIFT_MASK are set to 1.
• (modifiers & ~KeyEvent.SHIFT_MASK)-> every 1-bit in modifiersthat "does not belong" to KeyEvent.SHIFT_MASK
• if ((modifiers & ~KeyEvent.SHIFT_MASK) != 0)-> if there was at least one other bit set to 1 besides KeyEvent.SHIFT_MASKdo something...

• ~KeyEvent.SHIFT_MASK-> 除 KeyEvent.SHIFT_MASK 中的位之外的所有位都设置为 1。
• (modifiers & ~KeyEvent.SHIFT_MASK)-> modifiers“不属于”的每一位KeyEvent.SHIFT_MASK
• if ((modifiers & ~KeyEvent.SHIFT_MASK) != 0)-> 如果除了KeyEvent.SHIFT_MASK做某事之外，至少还有一个其他位设置为 1 ...

回答by Dhwaneel

From Java's website http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

从 Java 的网站http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

The unary bitwise complement operator "~" inverts a bit pattern; it can be applied to any of the integral types, making every "0" a "1" and every "1" a "0". For example, a byte contains 8 bits; applying this operator to a value whose bit pattern is "00000000" would change its pattern to "11111111".

一元按位补码运算符“~”反转位模式；它可以应用于任何整数类型，使每个“0”成为“1”，每个“1”成为“0”。例如，一个字节包含 8 位；将此运算符应用于位模式为“00000000”的值会将其模式更改为“11111111”。

Now, as previously answered by Pascal MARTIN, at any given case the value equals to -(x)-1. E.g. ~2=-3, ~-6=5, etc.

现在，正如 Pascal MARTIN 先前回答的那样，在任何给定情况下，该值都等于 -(x)-1。例如~2=-3，~-6=5，等等。

Also, in java all positive integersare stored as their binary representations and negative integersare stored in 2's compliment value of a positive integer.

此外，在 java 中，所有正整数都存储为它们的二进制表示，负整数存储在正整数的 2 的补码值中。

Now, let's see how it works in bit level in case of ~2=-3:

现在，让我们看看在 ~2=-3 的情况下它是如何在位级工作的：

Initially, 2 is stored in its binary representation:

最初，2 以其二进制表示形式存储：

0000 0000 0000 0010

Now ~2 will result in the value (inverse the bits):

现在 ~2 将导致值（反转位）：

1111 1111 1111 1101

How in the world I know it is -3? Well, it is -3 because it is derived from 2's compliment representation of 3.

我怎么知道它是-3？嗯，它是 -3，因为它是从 3 的 2 的恭维表示派生出来的。

As we know 2's(x)= 1's(x) + 1 (https://delightlylinux.wordpress.com/2014/10/13/binary-lesson-12-ones-complement-and-twos-complement/)
Our aim is it to find x:
1's(x)= 2's(x) - 1 (based on previous expression)

我们知道 2's(x)= 1's(x) + 1 ( https://delightlylinux.wordpress.com/2014/10/13/binary-lesson-12-ones-complement-and-twos-complement/)
我们的目标是否要找到 x:
1's(x)= 2's(x) - 1（基于前面的表达式）

As our answer is in is in 2's compliment,
1's(x)= 1111 1111 1111 1101 - 0000 0000 0000 0001
1's (x)= 1111 1111 1111 1100(How to subtract -http://sandbox.mc.edu/~bennet/cs110/pm/sub.html)

Therefore x= 1's compliment of value (as the answer we got represents 1's compliment of x).
x = 0000 0000 0000 0011
So, we have found that x is 3 and hence our previous result of ~ operator 1111 1111 1111 1101is -3 written as 2's compliment of 3.

由于我们的答案是 2 的恭维，
1's(x)= 1111 1111 1111 1101 - 0000 0000 0000 0001
1's (x)= 1111 1111 1111 1100（如何减去 - http://sandbox.mc.edu/~bennet/cs110/pm/sub.html）

因此 x= 1价值的赞美（因为我们得到的答案代表 1 对 x 的赞美）。
x =0000 0000 0000 0011
所以，我们发现 x 是 3，因此我们之前的 ~ 运算符的结果1111 1111 1111 1101是 -3 写成 2 的 3 的补码。