Help While Loop in C

Question

I'm new to C programming, I come from a Java background. I was wondering why in the following code, in the while loop I have to type my input ten times and then all ten inputs are displayed. I'm trying to type something once and have it displayed right after. Then continue typing my other inputs.

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


main(){

print(PROGRAM_INFO); //prints program name and author
print(PROMPT);

char input [100]; //array to hold input from user
int isActive = 1; //1 continue shell, 0 terminate shell
int count = 0;

while (isActive == 1 && count < 10){
    print(PROMPT);
    ++count;
    scanf("%s", input);
    print(input);


}


}

Mitch Wheat · Accepted Answer · 2009-09-20 03:51:56Z

5

Try flushing fflush(stdout) after each print(input)

answered Sep 20, 2009 at 3:51

Mitch Wheat

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2 Comments

LJM Over a year ago

yep, fflush would definitely be a better way to do it as compared to '\n'.

Sankofa Over a year ago

Thank you You helped me on a similar problem and now I don't have to ask it!

LJM · Accepted Answer · 2009-09-20 03:51:28Z

1

Try putting a '\n' character in the print. The stream just isn't being flushed.

answered Sep 20, 2009 at 3:51

LJM

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Comments

John Boker · Accepted Answer · 2009-09-20 03:51:35Z

1

after your final print statement you could probably call fflush(stdout); to flush stdout to the screen.

answered Sep 20, 2009 at 3:51

John Boker

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Comments

derobert · Accepted Answer · 2009-09-20 03:51:36Z

1

You need to flush STDOUT. fflush(STDOUT) should do it, added after the print.

answered Sep 20, 2009 at 3:51

derobert

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Comments

sud03r · Accepted Answer · 2009-09-20 07:38:39Z

1

by default input/output is buffered, i.e the input and output bytes are stored in a byte array before being displayed to the stream.
BUFSIZ in systems is generally a multiple of 1024.Although printf is line buffered, the buffer is flushed automatically when a newline is encountered.
fflush(stdout) causes the buffered data to be flushed to output stream which in this case is stdout. you can control buffer handling using setvbuf() function

edited Sep 20, 2009 at 7:38

answered Sep 20, 2009 at 4:14

sud03r

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3 Comments

Thomas Padron-McCarthy Over a year ago

Note that stdout is line buffered, so you don't need to wait for BUFSIZ characters.

William Pursell Over a year ago

fflush does not cause the buffered data to "be displayed on screen". It causes the stdout stream of the program to flush its buffers. If stdout is associated with a tty that is currently visible on the screen, then it incidentally causes text to be displayed on screen. (In other words: stdout is NOT the screen. Don't make that assumption.)

sud03r Over a year ago

@William,thomas.. thanks for your comments edited as pointed out.

pmg · Accepted Answer · 2009-09-20 13:35:20Z

0

Your print is a copy/paste error, right? It should be printf.

And you really shouldn't print the user string directly in the format of printf.
Imagine the user types "%d%f%s\a%c" ...

The best thing to do is

printf("%s\n", input);

With the '\n' in the format, you don't need to fflush(stdout); because stdout is line buffered by default and the '\n' does one on its own.

Besides, if the user types "%d%f%s\a%c" that's what you get printed.

The best thing to do is

puts(input);

puts adds the '\n' to the output and has no problems with format strings.

answered Sep 20, 2009 at 13:35

pmg

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