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Home/ Questions/Q 5848757
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T12:54:15+00:00 2026-05-22T12:54:15+00:00

It seems that when I run the following code: #include <stdio.h> #include <stdlib.h> int

  • 0

It seems that when I run the following code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char* argv)
{
    int i=0;

    setvbuf(stdout, NULL, _IOLBF,0);

    while (1)
            printf("%d ",i++);

    return 0;
}

it prints in chunks of 1024 chars, no matter the size I define for setvbuf().
The question is is if size affects somehow in this case and where is the definition for 1024 chars is coming from.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T12:54:16+00:00Added an answer on May 22, 2026 at 12:54 pm

    I don’t know how you identified 1024 but it’s probably BUFSIZ. BUFSIZ is defined in stdio.h.

    If buf is NULL, then the stdio library
    automatically allocates a buffer for
    use with stream (unless we select
    unbuffered I/O).

    EDIT

    Here’s something glibc says:

    Macro: int BUFSIZ
    The value of this macro is an integer constant expression that is
    good to use for the size argument to
    setvbuf. This value is guaranteed to
    be at least 256.

    The value of BUFSIZ is chosen on each system so as to make stream I/O
    efficient. So it is a good idea to use
    BUFSIZ as the size for the buffer when
    you call setvbuf.

    EDIT 2

    @larsmans is right. I looked at how setvbuf is implemented and it ignores a call when asking for line buffering and presenting a NULL buf. Now, stdout is no ordinary file, it’s attached to a terminal. So, heading over to pixelbeat

    • Buffer size only directly affects buffered mode
    • The default size like the kernel is based on the page size (4096 bytes on
      my system)
    • if stdin/stdout are connected to a terminal then default size = 1024;
      else size = 4096
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