Golang与python线程详解及简单实例
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 intvar cccc intvar m *sync.Mutexfunc 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 = 0def sumCount(temp): global count for i in range(temp): count = count + 1li = []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 = 0def 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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