swift中闭包(closure)

2016-05-27  本文已影响44人  玻璃瓶外的水

下面是一个基本的闭包:

var multiplyClosure = { (a: Int, b: Int) -> Int in

return a * b

}

这就是一个基本的闭包类型,其基本格式为{ (参数列表)-> 返回值 in 函数体}

对上面这个闭包,可以进行如下的精减

1、去掉return

var multiplyClosure = { (a: Int, b: Int) -> Int in

a * b

}

2、去掉返回值类型,让系统自己推断

var multiplyClosure = { (a: Int, b: Int) in

a * b

}

3、将参数中的类型声明去掉,这种情况要实现声明变量的类型为闭包

var multiplyClosure: (Int,Int) -> Int

multiplyClosure = { (a, b) in

a * b

}

4、最后还可以将参数列表去掉,根据参数出现的位置来处理

var multiplyClosure: (Int,Int) -> Int

multiplyClosure = {

$0 * $1

}

声明一个没有参数和返回值的闭包

let voidClosure: () -> Void = {

print("Swift is awesome")

}

voidClosure()

下面这个是闭包作为返回值的一个例子

func countingClosure() -> (() -> Int) {

print("ok")

var counter = 0

let incrementCounter: () -> Int = {

print(counter)

counter += 1

print(counter)

return counter

}

return incrementCounter

}

let counter1 = countingClosure()

let counter2 = countingClosure()

//counter1时一个闭包,闭包作为函数执行,返回int类型的值

counter1()

counter1()

counter2()

一个例子:

func mathSum(times: Int, operation: (Int) -> Int) -> Int {

var result = 0

for i in 1...times {

result += operation(i)

}

return result

}

var result1 = mathSum(10, operation: {(a: Int) -> Int in return a * a })

print(result1)

var result2 = mathSum(10, operation: {(a: Int) -> Int in fibonacci(a)})

print(result2)

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