C中的并行编程不执行指令

Question

I need help with C parallel programming, from this graph: graph image

I wrote this code:

#include <stdio.h>
#include <stdlib.h>
int main ()
{
    int r2;
    printf("---------   Program start   ---------");
    printf("\nBlock A instructions"); //block A
    printf("\nBlock B instructions"); //block B
    r2= fork();
    if (r2==0){ //child
        printf("\nBlock E instructions"); //block E
        printf("\nBlock F instructions"); //block F
        exit(0);
    }
    else{
        if (r2>0){ //father
            printf("\nBlock C instructions"); //block C
            printf("\nBlock D instructions"); //block D
            exit(0);
        }
        else printf("\nError");
    }
    printf("\nBlock G instructions"); //block G
    printf("\nBlock H instructions"); //block H
    printf("\n---------   Program finish   ---------");
}

And the output is:

---------   Program start   ---------
Block A instructions
Block B instructions
Block E instructions
Block F instructionsBlock B instructions
Block C instructions
Block D instructions

Why doesn't the program write the other instructions, and why does it write "block B instructions" twice?

--------------------- EDIT: ---------------------

The new code is:

#include <stdio.h>
#include <stdlib.h>
#include <sys/wait.h>
#include <unistd.h>
int main ()
{
    printf("---------   Program start   ---------\n");
    printf("Block A instructions\n"); //block A
    printf("Block B instructions\n"); //block B
    fflush(stdout);
    pid_t r2= fork();
    if (r2==0){ //child
        printf("Block E instructions\n"); //block E
        printf("Block F instructions\n"); //block F
        fflush(stdout);
        exit(1);
    }
    else{
        if (r2>0){ //father
            printf("Block C instructions\n"); //block C
            printf("Block D instructions\n"); //block D
            fflush(stdout);
            wait(NULL); //wait for child process to join with this parent, returns no value
        }
        else printf("Error");
    }
    printf("Block G instructions\n"); //block G
    printf("Block H instructions\n"); //block H
    printf("---------   Program finish   ---------");
    return 0;
}

The output now is:

---------   Program start   ---------
Block A instructions
Block B instructions
Block E instructions
Block F instructions
Block C instructions
Block D instructions
Block G instructions
Block H instructions
---------   Program finish   ---------

Sometimes the C and D instructions are written before the E and F instructions, is it normal or should it always be EF --> CD ? Btw is the code good or are there some mistakes? I compiled it with -Wall and I don't get any error messages

Answer 1

<stdio.h> is buffered , see stdio(3) and setvbuf(3) . You should call fflush(3) at appropriate places, in particular before fork(2) .

BTW, the code of your standard libc is free software , probably GNU glibc . Of course it uses syscalls(2) . So you should study its source code.

Read alsoHow to debug small programs

printf("\\nBlock C instructions");

is error-prone. The \\n could flush the buffer, and should in practice be used as the last character of printf(3) control format string.

Answer 2

It doesn't print the other instructions because of the exit(0) , so it will never reach G and H blocks. For the B being printed twice see Basile Starynkevitch's answer.