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我有三个进程，一个父亲和两个儿子，因为儿子们继承了在我分叉之前创建的所有东西我如何访问我之前从儿子或父亲那里创建的字符串数组？我想访问我在使用 string_hashes 的名称和来自儿子和父亲的哈希之前创建的两个字符串数组？谢谢您的帮助...

#include <unistd.h>
#include <sys/wait.h>
#include <sys/types.h>
#include <stdio.h>
#include <strings.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <errno.h>




#define PIPE_NAME   "np_workmaster"
#define max_chars_string 1000
#define n_childs 2

typedef struct
{
  int a;
  int b;
} numbers;



pid_t childs[n_childs];

void read_lines(char * filename, char (*pointer)[max_chars_string],int init_read,int n_lines);
void get_strings_hash(char (*pointer_strings)[max_chars_string],char (*pointer_hashes)[max_chars_string],int total_lines);

void worker(){ // meter as funcoes de ler passwords no filho
    char * ant = "";
    char hash_char;

    printf("[%d] I'm the son!\n", getpid());
    printf("[%d] My parent is: %d\n", getpid(), getppid());
    exit(0);
}

void pparent(){// sera o ultimo processo a terminar mostrando as passwords ja recebidas do dispatcher que recebe dos outros filhos
    printf("[%d] I'm the father!\n", getpid());
    //dispatcher() // criar funcao dispatcher que envia dados ao outro processo...

  int fd;
  if ((fd=open(PIPE_NAME, O_WRONLY)) < 0)
  {
    perror("Cannot open pipe for writing: ");
    exit(0);
  }

  // Do some work  
  while (1) {
       // here i  want to access arrays strings_hashes and hashes from the father and sons

    printf("[CLIENT] Sending (%d,%d) for adding\n", n.a, n.b);
    write(fd, &n, sizeof(numbers));
    sleep(2);
  }

  return 0;

}






int main(int argc, char **argv)
{
    char *filename;
    int status;//status do processos filho
    int resources[2];// numero de linhas que cada processo tem de ler
    //char * chave_iniciadora = "";
    int n_lines; //numero de linhas do ficheiro
    int i = 0;



    filename = (char*)malloc(strlen(argv[1])*sizeof(char)+1);

    if(argc !=3){
        fprintf(stderr, "Usage : %s [text_file] [cores]",argv[0]);
        exit(0);
    }

    strcpy(filename,argv[1]);
    n_lines = count_lines(filename); // contem o numero de linhas
    //definicao arrays
    char strings_hashes[n_lines][max_chars_string];//aray de string com as strings lidas do ficheiro
    char hashes[n_lines][max_chars_string]; // array de strings com as hashes
    char * pointer_string = &strings_hashes[0][0]; // ponteiro para o inicio do array das strings lidas do ficheiro
    char * pointer_hashes = &hashes[0][0];//ponteiro para o inicio do array das hashes

    //funcoes
    share_resources(atoi(argv[2]),n_lines,resources);
    read_lines(filename,strings_hashes,0,n_lines); // le as strings do ficheiro e passa para o array
    get_strings_hash(strings_hashes,hashes,n_lines);
     //
    for(i = 0; i<n_lines;i++){
        printf("%s",strings_hashes[i]);
    }
    for(i = 0; i<n_lines;i++){
        printf("%s",hashes[i]);
    }

    //   
  for(i = 0; i <atoi(argv[2]);i++){
        childs[i] = fork();
        if(childs[i] == -1){
           perror("Failed to fork");
           return 1;
       }
       if (childs[i] == 0)
       {
          worker();
       }
       else
       {
          pparent();
          wait(&status);
          if (!status)
            printf("\nOK\n");
          else 
            printf("\nSomething is wrong...\n");
    }
   }
  return 0;
}



///////////////////////////////////////////////////////////////////



//funcionar
void get_strings_hash(char (*pointer_strings)[max_chars_string],char (*pointer_hashes)[max_chars_string],int total_lines)//vai ao array de strings e corta a parte de hash e mete num array
{
    int i = 0;
    char *strings;
    char *hash;

    for(i = 0;i<total_lines;i++){
            strings = (char*)malloc(strlen(pointer_strings)*sizeof(char)+1);
            strcpy(strings,*pointer_strings);
            hash = (char*)malloc(strlen(pointer_strings)*sizeof(char)+1);
            find_hash(strings,hash);
            strcpy(*pointer_hashes,hash);
        pointer_hashes++;
        pointer_strings++;
    }

}



//funcionar
int count_lines(char * filename){ 
    FILE *fp;
    char str[max_chars_string];
    int i =0;

    if((fp = fopen("ficheiro_leitura.txt", "r"))==NULL) {
      printf("Cannot open file.\n");
      exit(1);
    }

    while(!feof(fp)) {
        while(fgets(str, sizeof str, fp)) {
           i++;
      }

  }
  fclose(fp);
  return i;
}
//funcionar
void read_lines(char * filename, char (*pointer)[max_chars_string],int init_read,int n_lines){ 
    FILE *fp;
    char str[max_chars_string];
    int i =0;


    if((fp = fopen(filename, "r"))==NULL) {
      printf("Cannot open file.\n");
      exit(1);
    }

    if(init_read>0 && init_read<=n_lines){
     for(i = 0;i<init_read;i++){
         fgets(str, sizeof str, fp);
       for(i = init_read;i<n_lines;i++){
           fgets(str, sizeof str, fp);
           strcpy(*pointer, str); //copia para a posicao actula do ponteiro
           pointer++;
       }
     }
    }
    if(init_read<=n_lines && init_read==0){
       for(i = init_read;i<n_lines;i++){
            fgets(str, sizeof str, fp);
           strcpy(*pointer, str); //copia para a posicao actula do ponteiro
           pointer++;
       }
     }


  fclose(fp);
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1 回答 1

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你不必做任何特别的事情。目前，您已将它们定义在main(). 要么全局定义它们，要么将它们传递给您的worker()（即子）函数。

请记住，它们现在是同一事物的 3 个独立副本，因此如果一个进程更改它们，这些更改将不会反映在其他进程中。如果这是您的意图，那么将它们放在共享内存中并用互斥锁保护它们可能最容易。

于 2012-10-11T23:05:01.800 回答