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Golang与python线程详解及简单实例

更新时间:2020-04-30 04:24:01 作者:startmvc
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

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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Golang与python线程详解 python线程