简单的多线程互斥示例不正确

Question

我希望以随机顺序从0到4得到数字,但相反,我有一些不同步的混乱

我做错了什么？

#include <iostream>
#include <windows.h>
#include <process.h>

using namespace std;

void addQuery(void *v );

HANDLE ghMutex;

int main()
{
    HANDLE hs[5];
    ghMutex = CreateMutex( NULL, FALSE, NULL);         
    for(int i=0; i<5; ++i)
    {
        hs[i] = (HANDLE)_beginthread(addQuery, 0, (void *)&i);
        if (hs[i] == NULL) 
        {
            printf("error\n"); return -1;
        }
    }

    printf("WaitForMultipleObjects return: %d error: %d\n",
         (DWORD)WaitForMultipleObjects(5, hs, TRUE, INFINITE), GetLastError());


    return 0;
}

void addQuery(void *v )
{
    int t = *((int*)v);
    WaitForSingleObject(ghMutex, INFINITE);

    cout << t << endl;

    ReleaseMutex(ghMutex);
    _endthread();
}

Answer 1

您必须在锁内读取和写入共享变量.您正在锁定之外读取它,从而使锁定无关紧要.

但即使这还不够,因为你的共享变量是一个循环变量,你正在编写而没有保护锁.一个更好的例子会像这样运行:

#include <iostream>
#include <windows.h>
#include <process.h>

using namespace std;

void addQuery(void *v );

HANDLE ghMutex;
int counter = 0;

int main()
{
    HANDLE hs[5];
    ghMutex = CreateMutex( NULL, FALSE, NULL);         
    for(int i=0; i<5; ++i)
    {
        hs[i] = (HANDLE)_beginthread(addQuery, 0, NULL);
        if (hs[i] == NULL) 
        {
            printf("error\n"); return -1;
        }
    }

    printf("WaitForMultipleObjects return: %d error: %d\n",
         (DWORD)WaitForMultipleObjects(5, hs, TRUE, INFINITE), GetLastError());


    return 0;
}

void addQuery(void *v)
{
    WaitForSingleObject(ghMutex, INFINITE);

    cout << counter << endl;
    counter++;

    ReleaseMutex(ghMutex);
    _endthread();
}

如果可以,请使用临界区而不是互斥锁,因为它们使用起来更简单,效率更高.但它们具有相同的语义,因为它们只保护锁定块内的代码.

注意:Jerry指出了其他一些问题,但我专注于高级别的trheading和序列化问题.