Java 按位非运算符
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bitwise not operator
提问by Sawyer
Why bitwise operation (~0);
prints -1 ? In binary , not 0 should be 1 . why ?
为什么按位运算(~0);
打印 -1 ?在二进制中,不是 0 应该是 1。为什么 ?
采纳答案by polygenelubricants
You are actually quite close.
你实际上很接近。
In binary , not 0 should be 1
在二进制中,不是 0 应该是 1
Yes, this is absolutely correct when we're talking about one bit.
是的,当我们谈论一点时,这是绝对正确的。
HOWEVER, an int
whose value is 0 is actually 32 bits of all zeroes! ~
inverts all 32 zeroes to 32 ones.
然而,一个int
值为 0 的实际上是 32 位全零!~
将所有 32 个零反转为 32 个 1。
System.out.println(Integer.toBinaryString(~0));
// prints "11111111111111111111111111111111"
This is the two's complement representation of -1
.
这是 的二进制补码表示-1
。
Similarly:
相似地:
System.out.println(Integer.toBinaryString(~1));
// prints "11111111111111111111111111111110"
That is, for a 32-bit unsigned int
in two's complement representation, ~1 == -2
.
也就是说,对于以int
二进制补码表示的 32 位无符号数,~1 == -2
.
Further reading:
进一步阅读:
- Two's complement
- This is the system used by Java (among others) to represent signed numerical value in bits
- JLS 15.15.5 Bitwise complement operator
~
- "note that, in all cases,
~x
equals(-x)-1
"
- "note that, in all cases,
- 补码
- 这是 Java(以及其他)用来以位表示有符号数值的系统
- JLS 15.15.5 按位补码运算符
~
- “请注意,在所有情况下,都
~x
等于(-x)-1
”
- “请注意,在所有情况下,都
回答by Bombe
Because ~
is not binary inversion, it's bitwise inversion. Binary inversion would be !
and can (in Java) only be applied to boolean values.
因为~
不是二进制反转,而是按位反转。二进制反转将!
并且可以(在 Java 中)仅应用于布尔值。
回答by cletus
In standard binary encoding, 0 is all 0s, ~
is bitwise NOT. All 1s is (most often) -1 for signed integer types. So for a signed byte type:
在标准二进制编码中,0 全为 0,~
按位非。对于有符号整数类型,所有 1 都是(最常见的)-1。所以对于有符号字节类型:
0xFF = -1 // 1111 1111
0xFE = -2 // 1111 1110
...
0xF0 = -128 // 1000 0000
0x7F = 127 // 0111 1111
0x7E = 126 // 0111 1110
...
0x01 = 1 // 0000 0001
0x00 = 0 // 0000 0000
回答by Daniel Fath
It's binary inversion, and in second complement -1 is binary inversion of 0.
它是二进制反转,在第二个补码中 -1 是 0 的二进制反转。
回答by LaZe
What you are actually saying is ~0x00000000 and that results in 0xFFFFFFFF. For a (signed) int in java, that means -1.
你实际上是在说 ~0x00000000,结果是 0xFFFFFFFF。对于 java 中的(有符号的)int,这意味着 -1。
回答by N 1.1
~
is a bitwise operator.
~
是位运算符。
~0 = 1 which is -1 in 2's complement form
http://en.wikipedia.org/wiki/Two's_complement
http://en.wikipedia.org/wiki/Two's_complement
Some numbers in two's complement form and their bit-wise not ~
(just below them):
一些二进制补码形式的数字和它们的按位不~
(就在它们下面):
0 1 1 1 1 1 1 1 = 127
1 0 0 0 0 0 0 0 = ?1280 1 1 1 1 1 1 0 = 126
1 0 0 0 0 0 0 1 = ?1271 1 1 1 1 1 1 1 = ?1
0 0 0 0 0 0 0 0 = 01 1 1 1 1 1 1 0 = ?2
0 0 0 0 0 0 0 1 = 11 0 0 0 0 0 0 1 = ?127
0 1 1 1 1 1 1 0 = 1261 0 0 0 0 0 0 0 = ?128
0 1 1 1 1 1 1 1 = 127
0 1 1 1 1 1 1 1 = 127
1 0 0 0 0 0 0 0 = ?1280 1 1 1 1 1 1 0 = 126
1 0 0 0 0 0 0 1 = ?1271 1 1 1 1 1 1 1 = ?1
0 0 0 0 0 0 0 0 = 01 1 1 1 1 1 1 0 = ?2
0 0 0 0 0 0 0 1 = 11 0 0 0 0 0 0 1 = ?127
0 1 1 1 1 1 1 0 = 1261 0 0 0 0 0 0 0 = ?128
0 1 1 1 1 1 1 1 = 127
回答by kpower
0 here is not a bit. It is a byte (at least; or more) - 00000000. Using bitwise or we will have 11111111. It is -1 as signed integer...
0这里不是一点。它是一个字节(至少;或更多) - 00000000。使用按位或我们将有 11111111。它是 -1 作为有符号整数......
回答by YOU
For 32 bit signed integer
对于 32 位有符号整数
~00000000000000000000000000000000=11111111111111111111111111111111
(which is -1)
~00000000000000000000000000000000=11111111111111111111111111111111
(这是-1)
回答by nickf
You could imagine the first bit in a signed number to be -(2x -1) where x is the number of bits.
您可以将有符号数中的第一位想象为 -(2 x -1),其中 x 是位数。
So, given an 8-bit number, the value of each bit (in left to right order) is:
因此,给定一个 8 位数字,每个位的值(按从左到右的顺序)是:
-128 64 32 16 8 4 2 1
Now, in binary, 0 is obviously all 0s:
现在,在二进制中,0 显然全是 0:
-128 64 32 16 8 4 2 1
0 0 0 0 0 0 0 0 0 = 0
And when you do the bitwise not ~
each of these 0s becomes a 1:
当你按位执行时,不是~
每个 0 都变成 1:
-128 64 32 16 8 4 2 1
~0 1 1 1 1 1 1 1 1
= -128+64+32+16+8+4+2+1 == -1
This is also helpful in understanding overflow:
这也有助于理解溢出:
-128 64 32 16 8 4 2 1
126 0 1 1 1 1 1 1 0 = 126
+1 0 1 1 1 1 1 1 1 = 127
+1 1 0 0 0 0 0 0 0 = -128 overflow!
回答by Nacho
I think the real reason is that ~ is Two's Complement.
我认为真正的原因是 ~ 是 Two's Complement。
Javascript designates the character tilde, ~, for the two's complement, even though in most programming languages tilde represents a bit toggle for the one's complement.
Javascript 指定字符代字号 ~ 表示二进制补码,尽管在大多数编程语言中,代字号代表一个二进制补码的切换。