每天成长一小步,积累下来就是一大步。
在GO中,开启15个线程,每个线程把全局变量遍历增加100000次,因此预测结果是 15*100000=1500000.
var sum int var cccc int var m *sync.Mutex func Count1(i int, ch chan int) { for j := 0; j < 100000; j++ { cccc = cccc + 1 } ch <- cccc } func main() { m = new(sync.Mutex) ch := make(chan int, 15) for i := 0; i < 15; i++ { go Count1(i, ch) } for i := 0; i < 15; i++ { select { case msg := <-ch: fmt.Println(msg) } } }
但是最终的结果,406527
说明需要加锁。
func Count1(i int, ch chan int) { m.Lock() for j := 0; j < 100000; j++ { cccc = cccc + 1 } ch <- cccc m.Unlock() }
最终输出:1500000
python中:同样方式实现,也不行。
count = 0 def sumCount(temp): global count for i in range(temp): count = count + 1 li = [] for i in range(15): th = threading.Thread(target=sumCount, args=(1000000,)) th.start() li.append(th) for i in li: i.join() print(count)
输出结果:3004737
说明也需要加锁:
mutex = threading.Lock() count = 0 def sumCount(temp): global count mutex.acquire() for i in range(temp): count = count + 1 mutex.release() li = [] for i in range(15): th = threading.Thread(target=sumCount, args=(1000000,)) th.start() li.append(th) for i in li: i.join() print(count)
输出1500000
OK,加锁的小列子。
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