by smallnest

Go单线程运行也会有并发问题

一个Go大佬群中严肃的讨论了一个问题：Go程序单线程多goroutine访问一个map会遇到并发读写panic么？

答案是肯定的，因为出现了这个问题所以大家才在群中讨论。

为什么呢？因为单线程意味着并行单元只有一个(多线程也可能并行单元只有一个)，但是多goroutine意味着并发单元有多个，如果并发单元同时执行，即使是单线程，可能就会产生数据竞争的问题，除非这些goroutine是顺序执行的。

举一个例子哈：

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func TestCounter() {
    runtime.GOMAXPROCS(1)
	
	var counter int

	var wg sync.WaitGroup
	wg.Add(10)
	for i := 0; i < 10; i++ {
		i := i
		go func() {
			fmt.Printf("start task#%d, counter: %d\n", i, counter)
			for j := 0; j < 10_0000; j++ {
				counter++
			}
			fmt.Printf("end task#%d, counter: %d\n", i, counter)
			wg.Done()
		}()
	}

	wg.Wait()
	fmt.Println(counter)
}

这段测试代码是启动10个goroutine对计数器加一，每个goroutine负责加10万次。在我的MBP m1笔记本上，每次的结果都是100万，符合期望。如果你运行这段代码，会发现goroutine其实是一个一个串行执行的(9->0->1->2->3->4->5->6->7->8，当然可能在你的机器上不是这样的)，如果是串行执行，不会有并发问题：

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start task#9, counter: 0
end task#9, counter: 100000
start task#0, counter: 100000
end task#0, counter: 200000
start task#1, counter: 200000
end task#1, counter: 300000
start task#2, counter: 300000
end task#2, counter: 400000
start task#3, counter: 400000
end task#3, counter: 500000
start task#4, counter: 500000
end task#4, counter: 600000
start task#5, counter: 600000
end task#5, counter: 700000
start task#6, counter: 700000
end task#6, counter: 800000
start task#7, counter: 800000
end task#7, counter: 900000
start task#8, counter: 900000
end task#8, counter: 1000000
1000000

为了制造点紧张气氛，我将代码改写成下面这样子，将counter++三条指令明显写成三条语句，并在中间插入runtime.Gosched(),故意给其它goroutine的执行制造机会：

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func TestCounter2() {
    runtime.GOMAXPROCS(1)

	var counter int

	var wg sync.WaitGroup
	wg.Add(10)
	for i := 0; i < 10; i++ {
		i := i
		go func() {
			fmt.Printf("start task#%d, counter: %d\n", i, counter)
			for j := 0; j < 10_0000; j++ {
				temp := counter
				runtime.Gosched()
				temp = temp + 1
				counter = temp
			}
			fmt.Printf("end task#%d, counter: %d\n", i, counter)
			wg.Done()
		}()
	}

	wg.Wait()
	fmt.Println(counter)
}

运行这段代码，你就会明显看到数据不一致的效果，即使是单个线程运行goroutine,也出现了数据竞争的问题：

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start task#9, counter: 0
start task#0, counter: 0
start task#1, counter: 0
start task#2, counter: 0
start task#3, counter: 0
start task#4, counter: 0
start task#5, counter: 0
start task#6, counter: 0
start task#7, counter: 0
start task#8, counter: 0
end task#9, counter: 100000
end task#1, counter: 100000
end task#3, counter: 100000
end task#2, counter: 100000
end task#5, counter: 100000
end task#0, counter: 100000
end task#4, counter: 100000
end task#6, counter: 100000
end task#7, counter: 100000
end task#8, counter: 100000
100000

这个结果非常离谱，期望100万，最后只有10万。

因为单个线程运行多个goroutine会有数据竞争的问题，所以访问同一个map对象也有可能出现并发bug,比如下面的代码，10个goroutine并发的写同一个map:

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func TestMap() {
	var m = make(map[int]int)
	var wg sync.WaitGroup
	wg.Add(10)
	for i := 0; i < 10; i++ {
		i := i
		go func() {
			fmt.Printf("start map task#%d, m: %v\n", i, len(m))
			for j := 0; j < 10_0000; j++ {
				m[j] = i*10_0000 + j
			}
			fmt.Printf("end map task#%d, m: %v\n", i, len(m))
			wg.Done()
		}()
	}

	wg.Wait()
}

大概率会出现panic:

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start map task#9, m: 0
start map task#0, m: 49152
fatal error: concurrent map writes

goroutine 41 [running]:
main.TestMap.func1()
	/Users/chaoyuepan/study/single_thread/main.go:72 +0xcc
created by main.TestMap in goroutine 1
	/Users/chaoyuepan/study/single_thread/main.go:69 +0x4c

goroutine 1 [semacquire]:
sync.runtime_Semacquire(0x140000021a0?)
	/usr/local/go/src/runtime/sema.go:62 +0x2c
sync.(*WaitGroup).Wait(0x1400000e1d0)
	/usr/local/go/src/sync/waitgroup.go:116 +0x74
main.TestMap()
	/Users/chaoyuepan/study/single_thread/main.go:79 +0xb8
main.main()
	/Users/chaoyuepan/study/single_thread/main.go:15 +0x2c