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 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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