丢掉所有goroutine都睡着了-死锁！----— Google GO中的错误

Question

我想编写三个同时发送整数的并发go例程。现在，我的代码已正确编译，但是在第一次执行后，出现错误“抛出：所有goroutine都处于睡眠状态-死锁！”。我试图找到错误，但在代码逻辑中找不到任何错误。有人可以帮助我在我的代码中查找错误吗？我的代码如下。

package main

import "rand"

func Routine1(command12 chan int, response12 chan int, command13 chan int, response13 chan int) {

    // z12 is a variable which stores the value comming from channel 2 and z13 is a variable which stores the value comming from channel 3.

    z12 := 200

    z13 := 200

    m12 := false

    m13 := false

    y := 0

    for i := 0; i < 20; i++ {

        y = rand.Intn(100)

        // If y's value is not 0 then the value will be sent to routine 2 or 3 according to   prime or not.

        // If y's value is 0 then process state (the varibles used by it means z12, z13) and channel state will be saved.[routine 1 is initiator]

        if y == 0 {

            print(z12, "    z12 STATE SAVED\n")

            print(z13, "    z13 STATE SAVED\n")

            // Routine 1 is initiator,  it sends 0 to make other process to save the state.

            y = 0

            command12 <- y

            command13 <- y

            // Untill routine 2 and 3 does not send 0, process 1 is on channel saving state (it's process state is already saved).

            // When routine 1 recives 0 from both other processes, channel is saved and routine 1 retuns to it's common routine procedure.

            // When routine 1 recives 0 from any other processes, saving channel bettwen them is stopped.

            // m12, m13 is used to mark whether 0 recived or not.

Answer 1

我不知道您的问题的答案，但是中的switch语句Routine3看起来有问题，因为它包含两个相同的case语句（这使我感到奇怪，为什么6g不会抱怨此代码）。一些使您的代码更清晰的建议：正如Evan已经指出的那样，请尝试为变量提供更多描述性的名称。读取的代码if someConditionIsMet比容易理解if m23 == false。通过将通用部分分解为函数来干燥代码。删除无效代码，例如将布尔值设置为true，然后检查其是否为true或检查奇数是否等于零。考虑使用else代替if <condition> {...}; if <negated condition> {...}我建议尝试提出单元测试，以详尽地描述您的功能的预期行为。这不仅可以帮助您找到错误，还可以提高您的编码技能。根据我的经验，考虑到测试编写的代码通常比未经测试的代码更易于理解，维护和发展。快乐黑客：）