Swift 中的 &<< 和 << 运算符有什么区别？

Question

Swift 中的 &<< 和 << 运算符有什么区别？看起来他们 return 结果相同：

print(2 << 3) // 16
print(2 &<< 3) // 16

Answer 1

FixedWithInteger protocol defines the “masking left shift operator” &<<为

Returns the result of shifting a value’s binary representation the specified number of digits to the left, masking the shift amount to the type’s bit width.

Use the masking left shift operator (&<<) when you need to perform a shift and are sure that the shift amount is in the range 0..<lhs.bitWidth. Before shifting, the masking left shift operator masks the shift to this range. The shift is performed using this masked value.

因此，如果移位量大于或等于左操作数的位宽，结果可能会有所不同：示例：

print(2 << 70) // 0
print(2 &<< 70) // 128

这里的移位量 (70) 大于 Int 的位数 (64)，因此 2 << 70 的计算结果为零。在第二行中，数字向左移动 70 % 64 = 6 位。

还有一个类似的“掩码右移运算符”&>>。示例：

let x = Int8.min        // -128 = 0b10000000
print(Int8.min >> 8)    //   -1 = 0b11111111
print(Int8.min &>> 8)   // -128 = 0b10000000

这里的第一个结果是 -1，因为将 有符号 整数向右移动会用 符号位填充左侧的空位。 第二个结果是-128 因为移位量为零：8 % 8 = 0.

命名和预期用途也在 SR-6749 中描述：

The goal of the operator, however, is never to have this wrapping behavior happen — it's what you use when you have static knowledge that your shift amount is <= the bitWidth. By masking, you and the compiler can agree that there's no branch needed, so you get slightly faster code without zero branching and zero risk of undefined behavior (which a negative or too large shift would be).

The docs are confusing because they give an example I don't think anyone would ever intentionally write—relying on the wrapping behavior to use some out of bounds value for the shift amount.

所以使用掩码移位运算符可以提高性能。许多示例可以在 Swift 标准库的源代码中找到，例如 UTF8.swift.

Swift 中的 &<< 和 << 运算符有什么区别？

What's the difference between &<< and << operators in Swift?

bit-manipulation

operators

bit

swift