du4629 2019-01-24 01:17
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I'm not understanding why this doesn't work https://play.golang.org/p/_ALPii0pXV6 but this https://play.golang.org/p/vCOjAr-o54e works.

As I understand the goroutine asynchronously sends to value true to a and 12 to b. While in the main function, a is blocked, until it receives a value. Why is it that when I rearrange it to have b is blocked before a, it results in a deadlock?

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  • douchui1488 2019-01-24 01:32

    Go channels are unbuffered by default. That means that it cannot send on a channel until the receiver is reading the channel. This is actually the Go preferred mode. It's more efficient than buffered channels in most cases.

    What that means for your first code is that the goroutine cannot proceed to write to channel b until it completes the write to channel a. It cannot do that until the main goroutine reads a.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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