Thread: Formatted output question...

Why, in certain situations, does calling printf() not print the given string to the screen (stdout) until the program terminates. I ran into this with a printf() nested inside an infinite for loop; inside that for loop a child process was created, and the printf() only printed out the string when the child was terminated.

I was told to use:

fprintf(stderr, "string in here");

This did in fact work, but I want a better understanding of why the normal printf() doesn't work in this situation. I can post the code if needed.

fflush(stdout), or put a newline in the output.

Originally Posted by Lateralus

Why, in certain situations, does calling printf() not print the given string to the screen (stdout) until the program terminates. I ran into this with a printf() nested inside an infinite for loop; inside that for loop a child process was created, and the printf() only printed out the string when the child was terminated.

I was told to use:

fprintf(stderr, "string in here");

This did in fact work, but I want a better understanding of why the normal printf() doesn't work in this situation. I can post the code if needed.

The answer has to do with buffered vs. unbuffered I/O. When you write something to stdout (via a standard printf call for example) it does not get sent immediately to the screen but rather gets sent into a buffer where it waits. When the buffer gets full, everything in the buffer then gets sent to the screen in one big chunk which is more efficient than accessing the hardware repeatedly for each individual bit of output that needs to get displayed. The buffer can also be flushed (and thus output to screen) prematurely via a call to the fflush function as already demonstrated by Dave. I think the buffer is also flushed when the program ends.

This is the case for stdout which is one of these buffered streams. The stderr stream however is an unbuffered stream. This means that when you write a string to stderr, that string is output immediately and does not sit around in a buffer somewhere.

Explained very well. Thanks for the quick response.