迅速的案件通过

Question

斯威夫特是否会通过声明落空？例如,如果我做以下

var testVar = "hello"
var result = 0

switch(testVal)
{
case "one":
    result = 1
case "two":
    result = 1
default:
    result = 3
}

是否可以为案例"一"和案例"两个"执行相同的代码？

Answer 1

是.你可以这样做:

var testVal = "hello"
var result = 0

switch testVal {
case "one", "two":
    result = 1
default:
    result = 3
}

或者,您可以使用fallthrough关键字:

var testVal = "hello"
var result = 0

switch testVal {
case "one":
    fallthrough
case "two":
    result = 1
default:
    result = 3
}

Answer 2

var testVar = "hello"

switch(testVar) {

case "hello":

    println("hello match number 1")

    fallthrough

case "two":

    println("two in not hello however the above fallthrough automatically always picks the     case following whether there is a match or not! To me this is wrong")

default:

    println("Default")
}

Answer 3

case "one", "two":
    result = 1

没有中断声明,但案例更灵活.

附录:正如Analog File指出的那样,breakSwift中确实有声明.它们仍然可以在循环中使用,虽然在switch语句中是不必要的,除非你需要填充其他空的情况,因为不允许空的情况.例如:default: break.

Answer 4

这是您容易理解的示例：

let value = 0

switch value
{
case 0:
    print(0) // print 0
    fallthrough
case 1:
    print(1) // print 1
case 2:
    print(2) // Doesn't print
default:
    print("default")
}

结论：fallthrough当前一个fallthrough匹配或不匹配时，用于执行下一个情况（仅一个）。