从在 C Linux 中的分叉之前创建的子进程访问字符串数组

Question

我有三个进程，一个父亲和两个儿子，因为儿子们通过遗产获得在我分叉之前创建的所有内容 如何访问我之前从儿子或父亲那里创建的字符串数组？ 我想访问我在分叉之前创建的两个字符串数组，这些字符串的名称是 string_hashes 和儿子和父亲的哈希？ 感谢您的帮助...

#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: %dn", 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 addingn", 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("nOKn");
          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);

Answer 1

你不必做任何特别的事情。 目前，您已在 main() 中定义了它们。 全局定义它们或将它们传递给您的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，那么最简单的方法是将它们放在共享内存中并使用互斥锁保护它们。