在多线程C程序错误线程ID？ [英] Wrong thread IDs in a multithreaded C program?

问题描述

我是新的C到多线程，我有这个问题。我写了下面code：

I am new to multithreading in C and I had this question. I wrote the following code:

#include <stdio.h>
#include <pthread.h>
#include <unistd.h>

pthread_mutex_t m=PTHREAD_MUTEX_INITIALIZER;
pthread_attr_t attr;

void* test(void *a)
{
    int i=*((int *)a);
    printf("The thread %d has started.\n",i);
    pthread_mutex_lock(&m);
    sleep(1);
    printf("The thread %d has finished.\n",i);
    pthread_mutex_unlock(&m);
    pthread_exit(NULL);

}

int main()
{
    int i=0;
    pthread_attr_setdetachstate(&attr,PTHREAD_CREATE_JOINABLE);
    pthread_t thread[5];

    for (i=0;i<5;i++)
        pthread_create(&thread[i],&attr,test,&i);

    for (i=0;i<5;i++)
        pthread_join(thread[i],NULL);
    return 0;
}

我

为什么会像值：

Why do I get values like:

The thread 0 has started.
The thread 0 has started.
The thread 5 has started.
The thread 5 has started.
The thread 0 has started.
The thread 0 has finished.
The thread 0 has finished.
The thread 5 has finished.
The thread 5 has finished.
The thread 0 has finished.

或

The thread 1 has started.
The thread 2 has started.
The thread 5 has started.
The thread 4 has started.
The thread 0 has started.
The thread 1 has finished.
The thread 2 has finished.
The thread 5 has finished.
The thread 4 has finished.
The thread 0 has finished.

甚至是：

The thread 0 has started.
The thread 0 has started.
The thread 0 has started.
The thread 0 has started.
The thread 0 has started.
The thread 0 has finished.
The thread 0 has finished.
The thread 0 has finished.
The thread 0 has finished.
The thread 0 has finished.

等等，当我有望获得：

etc, when I expected to get:

The thread 0 has started.
The thread 1 has started.
The thread 2 has started.
The thread 3 has started.
The thread 4 has started.
The thread 0 has finished.
The thread 1 has finished.
The thread 2 has finished.
The thread 3 has finished.
The thread 4 has finished.

只有当我把 usleep（10）在 thread_create 做我得到一些正常的值。

Only when I put usleep(10) after thread_create do I get some "normal" values.

我编译和运行code此code ::在Unix块。

I compiled and run this code in Code::Blocks on Unix.

推荐答案

请注意，您在传递 I 的地址作为参数传递给你的线程：

Notice that you are passing in the address of i as a parameter to your threads:

pthread_create(&thread[i],&attr,test,&i);

这意味着，所有的线程会被读取相同的变量 I ，以确定他们是哪个线程。也就是说，所有五个线程将看同样的变量来确定自己的线程号。因此，当值 I 在为循环增值，所有的线程会感觉到自己的线程数更改为使用的新值 I 。这就是为什么你会看到，有时5上来如线程数，也说明了事实，你经常逃课号码或看到太多的重复。

This means that all of your threads will be reading the same variable i to determine which thread they are. That is, all five threads will look at the same variable to determine their thread number. Consequently, as the value of i increments in your for loop, all the threads will perceive their thread number changing to use the new value of i. This is why you're sometimes seeing 5 come up as the thread number, and also explains the fact that you're often skipping numbers or seeing far too many duplicates.

要解决这个问题，你就需要给每个线程自己的拷贝我。例如，你可以做这样的事情：

To fix this, you will need to give each thread their own copy of i. For example, you could do something like this:

int* myI = malloc(sizeof(int));
*myI = i;
pthread_create(&thread[i], &attr, test, myI);

，然后让线程终止之前释放指针：

And then have the threads free the pointer before terminating:

void* test(void *a)
{
    int i=*((int *)a);
    printf("The thread %d has started.\n",i);
    pthread_mutex_lock(&m);
    sleep(1);
    printf("The thread %d has finished.\n",i);
    pthread_mutex_unlock(&m);
    pthread_exit(NULL);
    free(a);
}

另外，你可以投 I 到无效* ，并通过在：

pthread_create(&thread[i],&attr,test, (void*)i);

如果你这样做，你将不得不线程投下他们的论点直接回 INT ，而不是为int * ：

If you do this, you would then have the threads cast their arguments directly back to int, not to int*:

void* test(void *a)
{
    int i = (int)a;
    printf("The thread %d has started.\n",i);
    pthread_mutex_lock(&m);
    sleep(1);
    printf("The thread %d has finished.\n",i);
    pthread_mutex_unlock(&m);
    pthread_exit(NULL);
}

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