英文:
Why does Go have a "bit clear (AND NOT)" operator?
问题
为什么Go语言有&^运算符，即"位清除（AND NOT）"运算符？
a &^ b和a & ^b之间有什么区别吗？
英文:
Why does Go have &^, the "bit clear (AND NOT)" operator?
Is there ever any difference between a &^ b and a & ^b?
答案1
得分: 14
有一个微妙的区别，使用显式位清除运算符处理字面值和无类型常量会更容易。
无类型整数的默认类型是int，所以像a := uint32(1) & ^1这样的表达式是非法的，因为^1会先被计算，它被计算为^int(1)，结果为-2。然而，a := uint32(1) &^ 1是合法的，因为这里的1会根据上下文被计算为uint32类型。
使用显式位清除可能还会带来一些性能上的提升，但我对此不太确定。
英文:
There's a subtle difference that makes dealing with literals and untyped constants easier with the explicit bit clear operator.
Untyped integers have their default type as int so something like a := uint32(1) & ^1 is illegal as ^1 is evaluated first and it's evaluated as ^int(1), which equals -2. a := uint32(1) &^ 1 is legal however as here 1 is evaluated as uint32, based on the context.
There could also be some performance gains in having an explicit bit clear, but I'm not too sure about that.
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