普通网友 2015-09-24 08:15
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golang在死锁检测中有奇怪的行为

package main

import (
    "log"
    "net/http"
)

func useless_func(address string) []byte {
    http.Get("https://www.google.com")
    return nil
}
func test_a(test_channel chan int) {
    test_channel <- 1
    return
}

func test() {
    test_channel := make(chan int)
    for i := 0; i < 10; i++ {
        go test_a(test_channel)
    }
    for {
        log.Println(<-test_channel)
    }
}
func main() {
    test()
}

this code would not break because of deadlock, I try this code under Linux 4.1.6-1 and 3.16.0-4 with go 1.5.1 amd64 and got same result. but if i delete useless_func or use go 1.4.3 or run this under windows,it would perform well. this is really weird, if anyone could explain this?

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1条回答 默认 最新

  • dosi8657 2015-09-25 02:23
    关注

    Dominik Honnef provides the answer in response to issue ##12734 for Go 1.5.1:

    dominikh: The issue really lies with using cgo (which net uses, ignoring the details). When using cgo, the Go deadlock detection cannot function properly, because C world might call Go functions at any time, so in theory no deadlock exists; we might just be waiting for an external function call indefinitely.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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