Output not displayed with usleep until a line break is given

Question

I'm trying to program a simple "typewriter" effect in C, where text appears one letter at a time with a delay. Here's the function I have:

#include <stdio.h>
#include <unistd.h>

void typestring(const char *str, useconds_t delay)
{
    while (*str) {
        putchar(*(str++));
        usleep(delay);
    }
}

The problem is the text doesn't actually appear until a \n is displayed. What am I doing wrong?

I have been cautioned so many times against the fflush() thingie, especially fflush(stdin) that I have quitted using fflush() altogether. — Rüppell's Vulture, Apr 22 '13 at 08:21
Can anyone give an alternative solution to this typewriter problem without using fflush()? — Rüppell's Vulture, Apr 22 '13 at 08:21
I don't know if possible in Windows, but in Linux you can use `` and set the terminal to non-buffering. Look at the link in my post — bash.d, Apr 22 '13 at 08:23

score 3 · Accepted Answer · answered Apr 22 '13 at 08:18

3

The output to stdout is buffered. Using \n you are forcing a flush. If you want to change this, you will need to change the settings of the terminal (for Linux look here) or use

void typestring(const char *str, useconds_t delay)
{
    while (*str) {
        putchar(*(str++));
        fflush(stdout);
        usleep(delay);
    }
}

answered Apr 22 '13 at 08:18

bash.d

13,029
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Don't you mean `stdout`? Anyways, once I made that change, it worked flawlessly. Thanks a lot! – flarn2006 Apr 22 '13 at 08:20
You should flush `stdout` not `stdin`. – Rohan Apr 22 '13 at 08:20
Not sure, but the wise ones here have cautioned me many times against using fflush(stdin). – Rüppell's Vulture Apr 22 '13 at 08:20

score 2 · Answer 2 · answered Apr 22 '13 at 08:18

2

Your output stream might have get buffered, '\n' flushes the buffer.

Try fflush(stdout after putchar(), as

while (*str) {
        putchar(*(str++));
        fflush(stdout);
        usleep(delay);
    }

answered Apr 22 '13 at 08:18

Rohan

52,392
12
90
87

score 1 · Answer 3 · answered Apr 22 '13 at 08:20

1

\n implicitly forces output device to flush buffered input. To flush you should explicitly use fflush:

 fflush(stdout);

answered Apr 22 '13 at 08:20

masoud

55,379
16
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score 0 · Answer 4 · answered Apr 22 '13 at 09:07

Output Stream is buffered that's why text doesn't actually appear until a \n is displayed '\n' flushes the output stream (hard flush) to do manually the same you can call this function [fflush(stdout)].

while (*str) {
        putchar(*(str++));
        fflush(stdout);
        usleep(delay);
    }

or you can use

while (*str) {
            printf("%c\n",*(str++));
            usleep(delay);
        }

Output not displayed with usleep until a line break is given

4 Answers4

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