使用c ++ 11的std :: thread，何时修改全局变量是线程安全的？

Question

Consider this source:

#include <string>
#include <iostream>
#include <thread>

using namespace std;

int *up;

void testf(){
    for(int i = 0; i < 1000; i++)
        for(int f = 0; f < 11; f++)
            up[i]++;
}

int main() {
    up = new int[1000];

    std::thread tt[7];

    for(int ts=0;ts<7;ts++) {
        tt[ts]=std::thread(testf);
    }

    for(int ts=0;ts<7;ts++) {
        tt[ts].join();
    }

    for(int i = 0; i < 1000; i++)
        cout << up[i];
    cout << endl;
    delete[] up;
    return 0;
}

I'm purposely writing to the same int array without any mutex. The for loop in testf() will increment all members of int up[1000] by 11, and we have 7 threads. So the output should be 77777777... (2000 Sevens)

But sometimes when I run the exe, I get a patch of numbers like this:

...7777777066676672756866667777777777777377777366667777777...

Why does this happen?

(to compile on linux: g++ -std=c++11 -pthread )

Answer 1

The reason is that "up[i]++;" is not a thread safe operation. It does basically this:

1. read value of up[i]
2. add one to the read value
3. write value of up[i]

With two threads what should happen:

• Thread1 1) read value of up[i] (3)
• Thread1 2) add one to the read value (4)

• Thread1 3) write value of up[i] (4)

• Thread2 1) read value of up[i] (4)

• Thread2 2) add one to the read value (5)
• Thread2 3) write value of up[i] (5)

What could happen:

• Thread1 1) read value of up[i] (3)

• Thread2 1) read value of up[i] (3)

• Thread1 2) add one to the read value (4)

• Thread1 3) write value of up[i] (4)
• Thread2 2) add one to the read value (4)
• Thread2 3) write value of up[i] (4)

So both threads write 4 to the array!

To solve this, you either need a mutex or an atomic increment operation on the array: http://baptiste-wicht.com/posts/2012/07/c11-concurrency-tutorial-part-4-atomic-type.html