在子进程中从管道读取时，read系统调用返回-1

Question

To learn how Pipe IPC mechanism works, I wrote a simple program that creates two child processes which share data using a pipe. The first child process has to read data from a file and pass it to the pipe.

Afterwards, the second child process has to read it, convert it to uppercase and write it to another file. The read system call in the second child process returns -1 when reading from the pipe. Also when I execute the program, in some cases printf in the first child does not print anything and in other cases printf in the second child does not print, too. Could you please point the mistakes in the program which are causing the problems?

int main(int args[], char * argv[]) {

    int fd[2];

    long length;
    char buff1[250];
    char buff2[250];

    FILE * fptr1;
    FILE * fptr2;

    pid_t A, B;
    pipe(fd);

    A = fork();

    if (A == -1) {
        printf("error in fork of A\n");
        exit(1);
    }
    if (A == 0) {
        fptr1 = fopen(argv[1], "r"); // program receives file names as argument

        if (fptr1 == NULL) {
            printf("Erro in file open1\n");
            exit(1);
        }

        fseek(fptr1, 0 L, SEEK_END);
        length = ftell(fptr1);
        fseek(fptr1, 0 L, SEEK_SET);

        close(fd[0]);

        fread(buff1, length, 1, fptr1);
        buff1[length] = '\0';
        printf("buff1 = %s", buff1);
        write(fd[1], buff1, length);

        fclose(fptr1);
        exit(0);
    } else {
        B = fork();
        if (B == -1) {
            printf("Error in forking child B");
            exit(1);
        }
        if (B == 0) {
            fptr2 = fopen(argv[2], "w");

            if (fptr2 == NULL) {
                printf("Error in file open2\n");
                exit(1);
            }

            close(fd[1]);
            int n = read(fd[0], buff2, length);
            printf("n = %d\n", n);
            upper_string(buff2); // converts characters to uppecase
            fwrite(buff2, 1, length, fptr2);
            fclose(fptr2);
        }
    }
    return 0;
}

Answer 1

There are few things to take into account here. First thing i would like to point is that you do not need to use two fork() calls. In that case you have three processes working in parallel (parent process and two child process, one per each fork() call).

One important point to take into account when you work with processes working in parallel is synchronism. In your code you are creating two processes. Parent process does not wait for any of its child, so it finishes its execution, and if child processes have not finished, they will become child of init process. But appart from that, you have the typical producer consumer problem. One of your child produce something and the other consume it, but how they work in parallel, consumer need to know that the product is ready to be consumed. So, in this case, i think the easiest way to do this job is to use just one fork(), so child become the producer and the parent process (the consumer) wait until its child finish the job.