Question

我试图在Bool中累积结果进行多次检查：

我的代码是：

 var validParams = login.characters.count > 4;
 validParams &= password.characters.count > 6;
 validParams &= ...
 // more validations
 if (validParams) {...}

但我收到错误Binary operator '&=' cannot be applied to two 'Bool' operands。

如何在非详细模式下完成此工作或重写此代码？

Answer 1

&=是bitwise operation，只能应用于整数类型。由于short circuiting，您所需要的&&=实际上并不存在。你需要写

validParams = validParams && ...

Answer 2

作为@JeremyP says，&=是in-place bitwise AND operator，它只对整数类型进行操作。但是，没有什么可以阻止您为就地逻辑AND操作定义自己的&&=运算符：

infix operator &&= : AssignmentPrecedence

func &&=(lhs: inout Bool, rhs: @autoclosure () -> Bool) {
    // although it looks like we're always evaluating rhs here, the expression "rhs()" is
    // actually getting wrapped in a closure, as the rhs of && is @autoclosure.
    lhs = lhs && rhs()
}

然后您可以像这样简单地使用它：

func someExpensiveFunc() -> Bool {
    // ...
    print("some expensive func called")
    return false
}

var b = false

b &&= someExpensiveFunc() // someExpensiveFunc() not called, as (false && x) == false.
print(b) // false

b = true

b &&= someExpensiveFunc() // someExpensiveFunc() called, as (true && x) == x
print(b) // false

正如您所看到的，因为我们将rhs: &&=参数设为@autoclosure参数，我们可以对lhs为{{{{{{{ 1}}。

二元运算符＆＃39;＆amp; =＆＃39;不能适用于两个Bool＆＃39;斯威夫特的操作数

2 个答案: