Why does printf output not appear right away when stepping through the code?

Question

I'm using Anjuta and gdb on Fedora 20 and created a C Makefile project. The code looks like this:

#include <stdio.h>

int main (void)
{
°   printf ("1");
°   printf ("2");
°   printf ("3");

    return (0);
}

° means I set a breakpoint at that position.

Now when I debug the code, there's no output while the current line is one of these printf-functions. Only when I exit main '123' appears in the terminal.

If I add \n to the second printf argument, then '12' appears as output when I move from breakpoint 2 to the 3rd one.

score 5 · Accepted Answer · answered Nov 03 '13 at 20:53

By default, stdout is line buffered when writing to a terminal, fully buffered when writing to any other type of stream. Since you're not printing any newlines, the output is being buffered. You can change the buffering mode with setbuf(), end each string with newline, or call fflush() when you want printing to take plac.

score 4 · Answer 2 · edited May 23 '17 at 12:13

4

This is because printf writes to stdout which happens to be buffered. For more details see here.

edited May 23 '17 at 12:13

Community

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answered Nov 03 '13 at 20:53

Udo Klein

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Its always buffered... not "happens to be" – texasbruce Nov 03 '13 at 20:54
1

Well, you can make it unbuffered by calling setbuf() – Udo Klein Nov 03 '13 at 20:55
1

@texasbruce stdout is _not_ always buffered. 1st, it is implementation defined. 2nd If buffered or not, the mode can be changed.. – chux - Reinstate Monica Nov 03 '13 at 21:02

score 2 · Answer 3 · answered Nov 03 '13 at 20:53

2

Add fflush(stdout) after each printf. Your output is small and remains in buffer until the progam exits.

answered Nov 03 '13 at 20:53

dnk

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Why does printf output not appear right away when stepping through the code?

3 Answers3

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