duancaoqin6683 2019-03-16 08:51 采纳率: 0%
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Given the following simple Go program:

ch := make(chan int)
go fmt.Println(<- ch)
ch <- 2

If I replace go fmt.Println(<- ch) with go func(){fmt.Println(<-ch)}(), it works well.

But with the original version I get:

fatal error: all goroutines are asleep - deadlock!


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  • dpnzf48660 2019-03-16 08:55

    As defined in the spec:

    The function value and parameters are evaluated as usual in the calling goroutine, but unlike with a regular call, program execution does not wait for the invoked function to complete. Instead, the function begins executing independently in a new goroutine. When the function terminates, its goroutine also terminates. If the function has any return values, they are discarded when the function completes.

    So the <-ch is evaluated. But it cannot since the channel is empty, hence it blocks until something writes to it. But there is nothing writes to it ever, hence it's a deadlock.

    When you wrap it in an anonymous function - the same happens, but for the anonymous function. And then its body is evaluated concurrently to the main goroutine.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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