C语言 如何让 C 中的多个线程处理二维数组的同一个 for 循环？

How can i have multiple threads in C working on the same for loop of a two-dimensional array?

提问by programmer

i have a program in C.

我有一个 C 程序。

I have created 3 threads with pthread_create and i have created a mutex in order to lock/unlock the critical regions.

我使用 pthread_create 创建了 3 个线程，并且我创建了一个互斥锁以锁定/解锁关键区域。

The 3nd argumentof pthread_createis a pointer to a function that the thread will execute.

该3ND参数的在pthread_create是一个指向该线程将执行的功能。

In the examples i've found in the Web, this function is always very simplee.g. prints the thread id or prints a message.

在我在网上找到的例子中，这个函数 总是非常简单，例如打印线程 ID 或打印一条消息。

What happens when the function that the thread shall execute contains a for loop?

当线程要执行的函数包含for 循环时会发生什么？

Cause in my program i would like each one of the threads to work with a two dimensional array.

因为在我的程序中，我希望每个线程都使用二维数组。

Each thread shall find the sum of a line of a two-dimensional array. e.g.

每个线程应找到二维数组的一行的总和。例如

Thread1shall calculate the sum of first line of the 2-dimensional array

Thread1计算二维数组第一行的总和

Thread2shall calculate the sum of the second line
Thread1shall calculate the sum of the 3nd line
Thread3shall calculate the sum of the 3nd line

线程2将计算的第二行的总和
线程1将计算3ND线的总和
Thread3应计算3ND线的总和

I don't care about the order of the threads, but i need every thread to pick one of the lines.

我不关心线程的顺序，但我需要每个线程都选择其中一行。

I have the following code that sumstwo cells in the two dimensional array.

我有以下代码对二维数组中的两个单元格求和。

The program:

该程序：

1. creates NTHREADS

    for(i=0; i < NTHREADS; i++)
       {
          pthread_create( &thread_id[i], NULL, CalculateSum, NULL );
       }
   

2. Each thread waits for the others to finish

   for(j=0; j < NTHREADS; j++)
      {
         pthread_join( thread_id[j], NULL);
      }
   

3. the function that each thread shall execute but for ONE lineof the array and NOTfor the WHOLE array is

   void *CalculateSum(void *dummyPtr)
   {
      pthread_mutex_lock( &mutex1 );
   
    int i,j,sum = 0;
   
     for( i = 0; i <= N-1; i++) {
           for( j = 0; j <= M-1; j++) {
                   sum = dimensional_array[i][j] + dimensional_array[i][j];
           }
            printf(" Sum = %d\n", sum);
           }
   
      counter++;
   
      pthread_mutex_unlock( &mutex1 );
   }
   

1. 创建 NTHREADS

    for(i=0; i < NTHREADS; i++)
       {
          pthread_create( &thread_id[i], NULL, CalculateSum, NULL );
       }
   

2. 每个线程等待其他线程完成

   for(j=0; j < NTHREADS; j++)
      {
         pthread_join( thread_id[j], NULL);
      }
   

3. 每个线程应执行但对于数组的一行而不是对于整个数组的函数是

   void *CalculateSum(void *dummyPtr)
   {
      pthread_mutex_lock( &mutex1 );
   
    int i,j,sum = 0;
   
     for( i = 0; i <= N-1; i++) {
           for( j = 0; j <= M-1; j++) {
                   sum = dimensional_array[i][j] + dimensional_array[i][j];
           }
            printf(" Sum = %d\n", sum);
           }
   
      counter++;
   
      pthread_mutex_unlock( &mutex1 );
   }
   

The whole program is as follows: The program does not have any compilation error.

整个程序如下： 程序没有任何编译错误。

In order to run it you shall do: gcc -pthread program.c

为了运行它，你应该这样做：gcc -pthread program.c

    //program.c

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

   #define NTHREADS 3
   void *CalculateSum(void *);
   pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
   int  counter = 0;

   #define N 10
   #define M 10

   int dimensional_array[N][M];

   main()
   {
      pthread_t thread_id[NTHREADS];
      int i, j;

      for (i = 0; i <= N - 1; i++ )
           for( j = 0; j <= M - 1; j++)
                   dimensional_array[i][j] = i;

      for(i=0; i < NTHREADS; i++)
      {
         pthread_create( &thread_id[i], NULL, CalculateSum, NULL );
      }

      for(j=0; j < NTHREADS; j++)
      {
         pthread_join( thread_id[j], NULL);
      }



      printf("Final counter value: %d\n", counter);

      //print ARRAY
      for (i = 0; i <= N-1; i++ ) {
           for( j = 0; j <= M-1; j++)
                   printf("%d\t",dimensional_array[i][j]);
           printf("\n");
           }
   }
   //Calculate
   void *CalculateSum(void *dummyPtr)
   {
      pthread_mutex_lock( &mutex1 );

    int i,j,sum = 0;

     for( i = 0; i <= N-1; i++) {
           for( j = 0; j <= M-1; j++) {
                   sum = dimensional_array[i][j] + dimensional_array[i][j];
           }
            printf(" Sum = %d\n", sum);
           }

      counter++;

      pthread_mutex_unlock( &mutex1 );
   }

So, i would like each thread to find the sum of a line but i'm confused, i don't know how to do that.

所以，我希望每个线程都能找到一行的总和，但我很困惑，我不知道该怎么做。

In my program every time a thread calls the Calculate function, all the sum of the lines are computed and not just one

在我的程序中，每次线程调用计算函数时，都会计算所有行的总和，而不仅仅是一个

[Caution:For simplicity i sum the first element with it's own,the point is to understand how those threads can all take place in that for loop]

[注意：为简单起见，我将第一个元素与它自己的元素相加，重点是了解这些线程如何在该 for 循环中发生]

I would be glad if someone could help me

如果有人可以帮助我，我会很高兴

Thanks, in advance

提前致谢

回答by dasblinkenlight

You should create an array of per-thread parameters, and pass these to the threads one-by-one. In your case a single pointer to intis sufficient: you pass to the thread its index threadindexfrom zero to NTHREADS, and the thread passes back the sum for rows such that row % NTHREADS == threadindex.

您应该创建一个每个线程参数的数组，并将这些参数一个一个地传递给线程。在您的情况下，单个指针int就足够了：您将其索引threadindex从零传递给线程NTHREADS，并且线程将行的总和传回，使得row % NTHREADS == threadindex.

Here is how your thread function looks:

这是您的线程函数的外观：

void *CalculateSum(void *args)
{
    int *argPtr = args;

    int i,j,sum = 0;
    int threadindex = *argPtr;

    for( i = 0; i <= N-1; i++) {
        if (i % NTHREADS != threadindex) continue;
        for( j = 0; j <= M-1; j++) {
            sum += dimensional_array[i][j];
        }
    }

    pthread_mutex_lock( &mutex1 ); Mutex must go here
    counter++;
    pthread_mutex_unlock( &mutex1 );
    // Pass the value back:
    *argPtr = sum;
}


main()
{
    pthread_t thread_id[NTHREADS];
    int thread_args[NTHREADS];
    int i, j;

    pthread_mutex_init(&mutex1, NULL);

    for (i = 0; i <= N - 1; i++ )
        for( j = 0; j <= M - 1; j++)
            dimensional_array[i][j] = i;

    for(i=0; i < NTHREADS; i++)
    {
        thread_args[i] = i;
        pthread_create( &thread_id[i], NULL, CalculateSum, &thread_args[i]);
    }

    int sum = 0;
    for(j=0; j < NTHREADS; j++)
    {
        pthread_join( thread_id[j], NULL);
        sum += thread_args[i];
    }

    printf("Final counter value: %d. Total: %d\n", counter, sum);
}

回答by Johnny Mopp

To calc the sum of one line (ignoring the thread stuff):

计算一行的总和（忽略线程内容）：

void *CalculateSum(void *dummyPtr)
{
    int j,sum = 0;

    int i = (int)dummyPtr;
    for( j = 0; j <= M-1; j++) {
        sum += dimensional_array[i][j];
    }
    printf(" Sum = %d\n", sum);

    pthread_mutex_lock( &mutex1 );
    counter++;
    pthread_mutex_unlock( &mutex1 );
}

And then create the thread like this:

然后像这样创建线程：

int line_number = 2;    // Or whatever line to print`enter code here`
pthread_create( &thread_id[i], NULL, CalculateSum, (void *)line_number );

EDIT: put "counter++" back in.

编辑：把“counter++”放回去。