context包的withtimout

这里讨论复用context的情况

package main

import (
	"context"
	"fmt"
	"time"
)

func main() {
	// 传递带超时的上下文告诉阻塞函数
	// 超时过后应该放弃它的工作。
	ctx, cancel := context.WithTimeout(context.Background(), 50*time.Millisecond)
	defer cancel()

	f := func(ctx context.Context, tsMill int) {
		select {
		case <-time.After(time.Duration(tsMill) * time.Millisecond):
			fmt.Println("func timeout", tsMill)
		case <-ctx.Done():
			fmt.Println(ctx.Err()) // 打印"context deadline exceeded"
		}
	}

	f(ctx, 25)
	f(ctx, 30)

	fmt.Println("main end")
}

package main

import (
	"context"
	"fmt"
	"time"
)

func f1(ctx context.Context) {
	t := time.NewTicker(time.Duration(time.Second * 2))
	select {
	case <-t.C:
		fmt.Println("f1 tick", time.Now())
		return
	case <-ctx.Done():
		fmt.Println("f1 ctx.Done")
	}
}

func f2(ctx context.Context) {
	t := time.NewTicker(time.Duration(time.Second * 3))
	select {
	case <-t.C:
		fmt.Println("f2 tick", time.Now())
		return
	case <-ctx.Done():
		fmt.Println("f2 ctx.Done", time.Now())
	}
}

func main() {

	fmt.Println("main begin", time.Now())
	ctx, cancel := context.WithTimeout(context.Background(), time.Second*4)
	defer func() {
		cancel()
		fmt.Println("cancel()", time.Now())
	}()

	f1(ctx)
	f2(ctx)
	fmt.Println("main end", time.Now())
}

结果

[root@localhost d0831]# go run main.go 
main begin 2021-09-01 07:43:23.27511909 +0800 CST m=+0.000206196
f1 tick 2021-09-01 07:43:25.276065643 +0800 CST m=+2.001152697
f2 ctx.Done 2021-09-01 07:43:27.276073332 +0800 CST m=+4.001160388
main end 2021-09-01 07:43:27.276106832 +0800 CST m=+4.001193875
cancel() 2021-09-01 07:43:27.276113528 +0800 CST m=+4.001200561

可以看到，复用timeout的ctx是会减的，后者使用剩余的时间