按位异或 (^)
按位 XOR (^
) 运算符返回一个数字或 BigInt,其二进制表示形式在每个位位置都有 1
,其中任一操作数(但不是两个操作数)的对应位都是 1
。
¥The bitwise XOR (^
) operator returns a number or BigInt whose binary representation has a 1
in each bit position for which the corresponding bits of either but not both operands are 1
.
Try it
语法
描述
¥Description
^
运算符针对两种类型的操作数进行重载:编号和 BigInt。对于数字,该运算符返回一个 32 位整数。对于 BigInts,该运算符返回一个 BigInt。它首先 将两个操作数强制转换为数值 并测试它们的类型。如果两个操作数都变成 BigInt,则执行 BigInt XOR;否则,它将两个操作数都转换为 32 位整数 并执行数字按位异或。如果一个操作数变为 BigInt 而另一个操作数变为数字,则抛出 TypeError
。
¥The ^
operator is overloaded for two types of operands: number and BigInt. For numbers, the operator returns a 32-bit integer. For BigInts, the operator returns a BigInt. It first coerces both operands to numeric values and tests the types of them. It performs BigInt XOR if both operands become BigInts; otherwise, it converts both operands to 32-bit integers and performs number bitwise XOR. A TypeError
is thrown if one operand becomes a BigInt but the other becomes a number.
该运算符对 二进制补码 中操作数的位表示进行运算。第一个操作数中的每个位都与第二个操作数中的相应位配对:第一位到第一位,第二位到第二位,依此类推。该运算符应用于每对位,并按位构造结果。
¥The operator operates on the operands' bit representations in two's complement. Each bit in the first operand is paired with the corresponding bit in the second operand: first bit to first bit, second bit to second bit, and so on. The operator is applied to each pair of bits, and the result is constructed bitwise.
XOR 运算的真值表为:
¥The truth table for the XOR operation is:
x | y | x 异或 y |
---|---|---|
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
9 (base 10) = 00000000000000000000000000001001 (base 2) 14 (base 10) = 00000000000000000000000000001110 (base 2) -------------------------------- 14 ^ 9 (base 10) = 00000000000000000000000000000111 (base 2) = 7 (base 10)
超过 32 位的数字将丢弃其最高有效位。例如,以下超过 32 位的整数将被转换为 32 位整数:
¥Numbers with more than 32 bits get their most significant bits discarded. For example, the following integer with more than 32 bits will be converted to a 32-bit integer:
Before: 11100110111110100000000000000110000000000001 After: 10100000000000000110000000000001
对于 BigInts,没有截断。从概念上讲,将正 BigInt 理解为具有无限数量的前导 0
位,将负 BigInt 理解为具有无限数量的前导 1
位。
¥For BigInts, there's no truncation. Conceptually, understand positive BigInts as having an infinite number of leading 0
bits, and negative BigInts having an infinite number of leading 1
bits.
将任何数字 x
与 0
按位异或将返回转换为 32 位整数的 x
。不要使用 ^ 0
将数字截断为整数;使用 Math.trunc()
代替。
¥Bitwise XORing any number x
with 0
returns x
converted to a 32-bit integer. Do not use ^ 0
to truncate numbers to integers; use Math.trunc()
instead.
示例
使用按位异或
规范
Specification |
---|
ECMAScript Language Specification # prod-BitwiseXORExpression |
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也可以看看
¥See also