Golang, Python line Cheng ...

This is a creation in Article, where the information may have evolved or changed.

Growing one small step a day is a big step in the accumulation.

In go, 15 threads are opened, and each thread increments the global variable traverse by 100,000 times, so the predicted result is 15*100000=1500000.

var sum intvar cccc intvar m *sync. Mutexfunc Count1 (i int, ch Chan int) {for   J: = 0; J < 100000; J + + {      CCCC = CCCC + 1   }   ch <-cccc}fu NC Main () {   m = new (sync. Mutex)   ch: = Make (chan int, n) for   I: = 0; i <; i++ {      go Count1 (i, ch)   } for   I: = 0; i < 15 ; i++ {      Select {case      msg: = <-ch:         fmt. PRINTLN (MSG)}}}   

But the final result, 406527

Indicates that a lock needs to be added.

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: Not in the same way.

Count = 0def Sumcount (temp):    global count for    I in range (temp):        count = count + 1li = []for i in range]: 
  th = Threading. Thread (Target=sumcount, args= (1000000,))    Th.start ()    li.append (TH) for i in Li:    i.join () print (count)

Output results: 3004737

The instructions also need to be locked:

Mutex = Threading. Lock () Count = 0def Sumcount (temp):    Global Count    mutex.acquire () for    I in range (temp):        count = count + 1< C11/>mutex.release () Li = []for i in range]:    th = Threading. Thread (Target=sumcount, args= (1000000,))    Th.start ()    li.append (TH) for i in Li:    i.join () print (count)

Output 1500000

OK, lock the small.