goroutine使用了外面的变量，造成闭包问题。怎么解决呢？

Question

goroutine使用了外面的变量，造成闭包问题。怎么解决呢？

Answer 1

自己思考了一下。其实，既然是闭包。那么，只要将item作为参数，传递给这个循环goroutine，就可以了&nbsp;他作为形参，自然不会被修改// 从out队列接收数据&nbsp; &nbsp; &nbsp; &nbsp; for _,item := range result.Items {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; fmt.Println("got item: ",item)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; go func(rec interface{}) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; c.ItemChan <- rec&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }(item)&nbsp; &nbsp; &nbsp; &nbsp; }带缓冲的channel，我认为不是一个好的办法。因为，缓冲终归会满。

Answer 2

for _,item := range result.Items {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; //fmt.Println("got item: ",item)//把这个放 go func()里面&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; go func() {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; fmt.Println("got item: ",item)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; c.ItemChan <- item&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }()&nbsp; &nbsp; &nbsp; &nbsp; }这样你打印的就和接收到的数据是一样的了。因为go func 并不能保证立即执行，在它还没有执行的时候，item值已经被for循环改变了我有个疑问，你在写入的地方起goroutine是为了for循环不被写入阻塞吗?如果不想因为写入被阻塞，可以给channel定义一个缓冲。按照例子的，缓冲数定义为items的数量即可