Question

这里我有一个程序，其中父进程创建了几个子进程，传递每个都有一个独特的整数。然后每个子进程将读取的整数写回到父进程，该进程将结果打印到标准输出：

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/wait.h>
#define R 0
#define W 1

void run_child(int in, int out){
    int r;
    int it;
    while((r = read(in, &it, sizeof(it))) != 0){
        if(r == -1){
            perror("read");
            exit(1);
        }
        //word[r] = '\0';
        int w = write(out, &it, sizeof(it));
        if(w == -1){
            perror("write");
            exit(1);
        }
        if(close(out) == -1){
            perror("close");
            exit(1);
        }
    }
}
int main(int argc, char **argv) {

    // Process fan

    int i;
    int n;
    int num_kids;
    int from_parent[2];
    int to_parent[2];


    if(argc != 2) {
        fprintf(stderr, "Usage: fan_write <numkids>\n");
        exit(1);
    }

    num_kids = atoi(argv[1]);
    int status;

    char word[32];
    for(i = 0; i < num_kids; i++) {
        if(pipe(from_parent) == -1){
            perror("pipe");
            exit(1);
        }
        if(pipe(to_parent) == -1){
            perror("pipe");
            exit(1);
        }
        int g = i;
        write(from_parent[W], &g, sizeof(int));

        n = fork();
        if(n < 0) {
            perror("fork");
            exit(1);
        }
        if(n == 0){
            if(close(from_parent[W]) == -1){
                perror("close");
                exit(1);
            }
            if(close(to_parent[R]) == -1){
                perror("close");
                exit(1);
            }
            dup2(from_parent[R], STDIN_FILENO);
            if(close(from_parent[R]) == -1){
                perror("close");
                exit(1);
            }
            run_child(STDIN_FILENO, to_parent[W]);
            close(to_parent[W]);
            exit(0);
        }

        if(close(from_parent[R]) == -1){
            perror("close");
            exit(1);
        }
        if(close(to_parent[W]) == -1){
            perror("close");
            exit(1);
        }

        if(close(from_parent[W]) == -1){
            perror("close");
            exit(1);
        }

        for(i=0;i<num_kids;i++){
            int read_int;
            int r = read(to_parent[R], &read_int, sizeof(int));
            printf("read %d bytes\n", r);
            if(r == -1){
                perror("read");
                exit(1);
            }
            printf("%d\n", read_int);
        }
    }
    for(i = 0; i < num_kids; i++){
        wait(&status);
    }

    return 0;
}

使用num_kids = 4，我希望程序每次读取4个字节并打印出来不同的整数。但是，运行时，它在一次迭代中读取4个字节，然后读取在以下迭代中为0个字节，并反复打印相同的整数。我不确定如何解决它。

编辑：解决了！提示：对管道使用文件描述符矩阵。

Answer 1

你有错误的概念来读取循环中所有num_kids子进程中的数字，你每次都是从单个孩子和每个孩子中读取循环：

   for(i=0;i<num_kids;i++){ 
        int read_int;
        int r = read(to_parent[R], &read_int, sizeof(int));
        printf("read %d bytes\n", r);
        if(r == -1){
            perror("read");
            exit(1);
        }
        printf("%d\n", read_int);
    }

代码是否也为包括父项在内的每个子项运行，因为您无条件地在子/父进程中运行此循环。但是父母可以从单个孩子那里读到你第一次获得4个字节然后0的原因。因为孩子回来了一次。从您的代码中删除上面的for循环，并像我在下面建议的那样：

你应该这样做（阅读评论）：

if(n == 0){
   //child
}
else{ 
  // read in parent without loop
}

正确的方法是：

    if(n == 0){//child
        if(close(from_parent[W]) == -1){
            perror("close");
            exit(1);
        }
        if(close(to_parent[R]) == -1){
            perror("close");
            exit(1);
        }
        dup2(from_parent[R], STDIN_FILENO);
        if(close(from_parent[R]) == -1){
            perror("close");
            exit(1);
        }
        run_child(STDIN_FILENO, to_parent[W]);
        close(to_parent[W]);
        exit(0);
    }
    else{ // parent
       write(from_parent[W], &g, sizeof(int)); 
       int read_int;
       int r = read(to_parent[R], &read_int, sizeof(int));
        printf("read %d bytes\n", r);
        if(r == -1){
            perror("read");
            exit(1);
        }
        printf("%d\n", read_int);
    }

它的工作方式如下：

:~$ ./a.out  4
read 4 bytes
0
read 4 bytes
1
read 4 bytes
2
read 4 bytes
3

Answer 2

if (n == 0) { //child
    if(close(from_parent[W]) == -1) {
        perror("close");
        exit(1);
    }
    if(close(to_parent[R]) == -1) {
        perror("close");
        exit(1);
    }
    run_child(from_parent[R], to_parent[W]);
    close(from_parent[R]);   // ignore checking return code here!
    close(to_parent[W]);     // ignore checking return code here!
    exit(0);
}

// And this is what run_child looks like
void run_child(int in, int out){
    int r;
    int it;
    while((r = read(in, &it, sizeof(it))) != 0){
        if(r == -1){
            perror("read");
            exit(1);
        }

        int w = write(out, &it, sizeof(it));
        if(w == -1){
            perror("write");
            exit(1);
        }
    }
}

Answer 3

我把答案代码混淆了，这是最后的结果.. 一切顺利

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/wait.h>
#define R 0
#define W 1

void run_child(int in, int out){
    int r;
    int it;
    while((r = read(in, &it, sizeof(it))) != 0){
        if(r == -1){
            perror("read");
            exit(1);
        }

        int w = write(out, &it, sizeof(it));
        //perror("write:");
        //close(out) ;

        exit(1);

    }
}
int main(int argc, char **argv) {

    // Process fan

    int i;
    int n;
    int num_kids;
    int from_parent[2];
    int to_parent[2];


    if(argc != 2) {
        fprintf(stderr, "Usage: fan_write <numkids>\n");
        exit(1);
    }

    num_kids = atoi(argv[1]);
    int status;

    char word[32];
    for(i = 0; i < num_kids; i++) {
        if(pipe(from_parent) == -1){
            perror("pipe");
            exit(1);
        }
        if(pipe(to_parent) == -1){
            perror("pipe");
            exit(1);
        }
        int g = i;
        write(from_parent[W], &g, sizeof(int));

        n = fork();
        if(n < 0) {
            perror("fork");
            exit(1);
        }

        if(n == 0){//child
            if(close(from_parent[W]) == -1){
                perror("close");
                exit(1);
            }
            if(close(to_parent[R]) == -1){
                perror("close");
                exit(1);
            }
            run_child(from_parent[R], to_parent[W]);

            if(close(from_parent[R]) == -1){
                perror("close");
                exit(1);
            }
            if(close(to_parent[W]) == -1){
                perror("close");
                exit(1);
            }

            exit(0);
        }
        else{ // parent
           write(from_parent[W], &g, sizeof(int)); 
           int read_int;
           int r = read(to_parent[R], &read_int, sizeof(int));
            printf("read %d bytes\n", r);
            if(r == -1){
                perror("read");
                exit(1);
            }
            printf("%d\n", read_int);
        }

    }


    for(i = 0; i < num_kids; i++){
        wait(&status);
    }

    return 0;
}

多个孩子读/写单亲

3 个答案: