C fork程序说明

Question

I have a code as below

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main(int argc, char *argv[])
{
    printf("a\n");
    fork();

    printf("b\n");
    if(fork() == 0) {
        printf("c\n");
        exit(0);    
    }
    printf("d\n");
    return 0;
}

Output :

a
b
d
a
b
c
a
b
d
a
b
c

I don't know why the output duplicated many times.

Answer 1

I don't know why the output duplicated many times

Because printf() is buffered.

When a process calls fork() , the resulting child process obtains a copy of the parent's output buffer.

You can empty this output buffer by placing fflush(stdout) just before each call to fork() . In that case output should be:

a
b
b
d
c
d
c

---

Note that if the output refered to a terminal, it would be line-buffered by default, ie: the buffer would be dumped every time a \\n is sent to the output. Not so if you are redirecting the output to a file.

Answer 2

When you call fork() it gets a copy of output buffer of the calling process. Buffering is enabled by default, so you get this behavior. You can use

fflush(stdout);

before a call to fork(). Or, you can also disable buffering using

setbuf(stdout, NULL);

You can read more about fork here . Let me know if you need any more help.

Answer 3

The answer is already in the comments. You are calling fork() twice. So the solution is to just call it once and save the result in a variable like this int pid = fork() . Also, you should check if the fork-call failed (if it does, it returns a negative value).

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main(int argc, char *argv[])
{
    printf("a\n");

    int pid = fork();
    if (pid < 0)
    {
        fprintf(stderr, "Can't fork!");
        exit(1);
    }

    printf("b\n");

    if(pid == 0)
    {
        printf("c\n");
    }
    else
    {
        printf("d\n");
    }

    return 0;
}