break

  • select中的break,类似c系列中的break,break后的语句不执行
  • for和select一同使用,有坑
    break只能跳出select,无法跳出for
package test

import (
    "fmt"
    "testing"
    "time"
)

func TestBreak(t *testing.T) {
    tick := time.Tick(time.Second)
    for {
        select {
        case t := <-tick:
            fmt.Println(t)
            break
        }
    }
    fmt.Println("end")
}

执行结果:

=== RUN   TestBreak
2019-12-19 14:43:41.7912242 +0800 CST m=+1.005627701
2019-12-19 14:43:42.0862832 +0800 CST m=+1.007127901
2019-12-19 14:43:42.7914242 +0800 CST m=+2.005754701
2019-12-19 14:43:43.0864832 +0800 CST m=+2.007254901
...

break无法跳出select的解决方案

1、标签

func TestBreak(t *testing.T) {
    tick := time.Tick(time.Second)
//FOR是标签
FOR:
    for {
        select {
        case t := <-tick:
            fmt.Println(t)
            //break出FOR标签标识的代码
            break FOR
        }
    }
    fmt.Println("end")
}

2、goto


func TestBreak(t *testing.T) {
    tick := time.Tick(time.Second)
    for {
        select {
        case t := <-tick:
            fmt.Println(t)
            //跳到指定位置
            goto END
        }
    }
END:
    fmt.Println("end")
}

执行结果:

=== RUN   TestBreak
2019-12-19 14:43:41.7912242 +0800 CST m=+1.005627701
end

continue

单独在select中是不能使用continue,会编译错误,只能用在for-select中。
continue的语义就类似for中的语义,select后的代码不会被执行到。

func TestBreak(t *testing.T) {
    tick := time.Tick(time.Second)
    for {
        select {
        case t := <-tick:
            fmt.Println(t)
            continue
            fmt.Println("test")
        }
    }
    fmt.Println("end")
}

执行结果

=== RUN   TestBreak
2019-12-19 14:43:41.7912242 +0800 CST m=+1.005627701
2019-12-19 14:43:42.0862832 +0800 CST m=+1.007127901
2019-12-19 14:43:42.7914242 +0800 CST m=+2.005754701
2019-12-19 14:43:43.0864832 +0800 CST m=+2.007254901
...

return

和函数中的return一样,跳出select,和for,后续代码都不执行

12-19 11:25