按位与 (&)

¥Syntax

x & y

¥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为 BigInt，则执行 BigInt AND；否则，它将两个操作数都转换为 32 位整数 并执行数字按位与。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 AND if both operands become BigInts; otherwise, it converts both operands to 32-bit integers and performs number bitwise AND. 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.

AND 运算的真值表为：

¥The truth table for the AND operation is:

     9 (base 10) = 00000000000000000000000000001001 (base 2)
    14 (base 10) = 00000000000000000000000000001110 (base 2)
                   --------------------------------
14 & 9 (base 10) = 00000000000000000000000000001000 (base 2) = 8 (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.

对任何数字 x 与 -1 进行按位与运算将返回转换为 32 位整数的 x。不要使用 & -1 将数字截断为整数；使用 Math.trunc() 代替。

¥Bitwise ANDing any number x with -1 returns x converted to a 32-bit integer. Do not use & -1 to truncate numbers to integers; use Math.trunc() instead.

¥Examples

¥Using bitwise AND

// 9  (00000000000000000000000000001001)
// 14 (00000000000000000000000000001110)

14 & 9;
// 8  (00000000000000000000000000001000)

14n & 9n; // 8n

¥See also

• JS 指南中的位运算符
• 按位与赋值 (&=)