Golang and python thread details and simple examples

  • 2020-05-30 20:32:13
  • OfStack

Golang and python threads details and simple examples

In GO, 15 threads are started and each thread iterates the global variable 100,000 times, so the predicted result is 15* 100,000 = 1,500,000.


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)
   }
  }
}

But the end result, 406,527

It says it needs to be locked.


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

Final output: 1500000

python: same way, no.


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)

Output result: 3004737

Note also need to be locked:


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)

The output of 1500000

OK, small column with locks.

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