Golang、python線程小列子。。。。。。

天天成長一小步,積累下來就是一大步。python

在GO中,開啓15個線程,每一個線程把全局變量遍歷增長100000次,所以預測結果是 15*100000=1500000.app

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)
      }
   }
}

可是最終的結果,406527ui

說明須要加鎖。線程

func Count1(i int, ch chan int) {
   m.Lock()
   for j := 0; j < 100000; j++ {
      cccc = cccc + 1
   }
   ch <- cccc
   m.Unlock()
}

最終輸出:1500000get

 

python中:一樣方式實現,也不行。thread

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)

輸出結果:3004737變量

說明也須要加鎖:select

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)

輸出1500000遍歷

OK,加鎖的小列子。di

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