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How much XD will be printed?


I ran into this weird problem while playing with fork () with one of my friends. Very simple POC code:

int main(int argc, char** argv)
{
    int i = 0;
    for(i=0; i<4; i++) {
        printf("xd\n");
        fork();
    }
    return 0;
}

I got a good result: xd xd xd xd xd xd xd xd xd xd xd xd xd xd xd

xd was printed 15 times what I expected - the number of nodes in a 4-level full binary tree. But when we removed the "\ n" in printf, we got another completely different result:

xdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxdxd

This gave me 64 XD (my friend got 56 XD in his machine). This seems to be a somewhat stable result as I could run it many times and it gives me the same output (same as my friend).

I tried to change printf ("xd") to perror ("xd"), it gave me 15 outputs: xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success xd: Success

And I tried

    for(i=0; i<4; i++) {
        printf("xd");
        fflush(stdout);
        fork();
    }

15 XD .
, , .
, concurrency, , fork() , i. (, , )
, , stdout, , , , , - , / .
Both of these are probably wrong as I don't know much about Linux implementations, can anyone help me explain this?

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2 answers


The behavior you are doing is related to the buffering mode.

Since there is no newline at the end and the output is in line-buffered mode (or full-buffered mode), so the output you see is different



You should use fflush(0);

to remove all I / O buffers before fork

You can control the buffering mode with standard C functions setvbuf()

and _IOFBF

(full buffering), _IOLBF

(line buffering) and _IONBF

(no buffering).

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stdout has a buffering string, so in the first version everything printf

will be printed at once:

i=0: 1 xd
i=1: 2 xd
i=2: 4 xd
i=3: 8 xd
in sum: 15 xd

on the other hand without the \n

stdout buffer will be copied with each fork()

and thus doubles the xd count.



i=0:  1 xd, after fork  2 xd and  2 processes
i=1:  4 xd, after fork  8 xd and  4 processes
i=2: 12 xd, after fork 24 xd and  8 processes
i=3: 32 xd, after fork 64 xd and 16 processes

at the end of the main process all 64 xd's will be printed.

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