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