Question

我宣布了一个数组：

char * words[1000] = {NULL};

现在我有一系列分叉子进程向该数组添加单词，但它们不会影响父程序。我怎么能改变它？

Answer 1

你没有在while块中添加if块!!!

   while(i < 1000 && words[i] != NULL)
   {
   i++;
 if(i<1000){
    words[i] = (char*) malloc(strlen(temp)+1);
    strcpy(words[i], temp);
   words[i][strlen(words[i])] = '\0';
    printf("Added: %s at location %d\n", words[i], i);
   }
  }

Answer 2

试试这个你得到的支架错误：

int i = 0;
while(i < 1000 && words[i] != NULL){
    i++;
    if(i<1000){
        words[i] = (char*) malloc(strlen(temp)+1);
        strcpy(words[i], temp);
        words[i][strlen(words[i])] = '\0';
        printf("Added: %s at location %d\n", words[i], i);
    }
}

Answer 3

请粘贴整个代码......我编写代码来做我认为你正在做的事情，它对我有用......

    #include <stdio.h>

int main(int argc, char *argv[])
{
  char* words[1000];
  int j;
  for(j = 0; j<1000; j++)
    words[j] = NULL;
  char *temp = "dummy";

  for (j = 0; j < 10; j++)
    {
      int i = 0;
      while(i < 1000 && words[i] != NULL)
        i++;
      printf("Adding something to %d vs %d\n",i,j);
      if(i<1000){
        words[i] = (char*) malloc(strlen(temp)+1);
        strcpy(words[i], temp);
        words[i][strlen(words[i])] = '\0';
        printf("Added: %s at location %d\n", words[i], i);
      }
    }
}

/* prints:
Adding something to 0 vs 0
Added: dummy at location 0
Adding something to 1 vs 1
Added: dummy at location 1
Adding something to 2 vs 2
Added: dummy at location 2
Adding something to 3 vs 3
Added: dummy at location 3
Adding something to 4 vs 4
Added: dummy at location 4
Adding something to 5 vs 5
Added: dummy at location 5
Adding something to 6 vs 6
Added: dummy at location 6
Adding something to 7 vs 7
Added: dummy at location 7
Adding something to 8 vs 8
Added: dummy at location 8
Adding something to 9 vs 9
Added: dummy at location 9
*/

Answer 4

嗯，对于你的编辑案例：不要使用fork，使用线程，因为那样一切都在一个地址空间中运行...
当然，然后使用互斥锁保护你的单词阵列......

Answer 5

当您使用fork创建子进程时，每个子进程都会获得自己的数组副本，而当子进程更改数组中的某些内容时，它实际上正在更改自己的副本，对于您要执行的操作，您需要进程间通信（IPC）创建共享内存或创建管道，以便更改所有子代和父代的数组值

如何让子进程更改父进程的变量？

5 个答案: