echo charecter program...

[hdsalbki]

hi everyone,
I have below a small program to echo back what character the user
types. It's working OK but prints a extra prompt between every
character. what can I do? I'm new in C and my book is very
difficult...
thanks for any help.

#include<stdio. h>

main()
{
char a;
while(1)
{
printf("\nPleas e type a character: ");
scanf("%c",&a);
printf("\nYou typed: %c",a);
}
}

[santosh]

hdsalbki wrote:

CBFalconer wrote:

<snip>

>A handy routine to keep available is flushln, below. For your
application you could call "flushln(stdin) ;". You can also test it
for EOF.

int flushln(FILE *f) {
int ch;
while ((EOF != (ch = getc(f))) || (ch != '\n')) continue;
return ch;
} /* flushln */

for example:

do {
/* something that eats chars other than '\n' */
} while (EOF != flushln(stdin)) ;

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

Please don't quote sigs (the text that follows the '-- ' bit) unless you
are specifically commenting on it.

Thanks CBFalconer. But when I run the following program which uses
your code, I can't get past the first prompt. The app just seems to
there and all I can do is kill it with ctrl-c. Am I doing something
wrong? I copies and pasted your code so to avoid typo...

#include <stdio.h>

int flushln(FILE *f) {
int ch;
while ((EOF != (ch = getc(f))) || (ch != '\n')) continue;

This needs to use the && logical operator to function as Chuck probably
intended it:

while ((EOF != (ch = getc(f))) && (ch != '\n')) continue;

<snip rest>

[hdsalbki]

santosh wrote:

#include <stdio.h>

int flushln(FILE *f) {
int ch;
while ((EOF != (ch = getc(f))) || (ch != '\n')) continue;

This needs to use the && logical operator to function as Chuck probably
intended it:

while ((EOF != (ch = getc(f))) && (ch != '\n')) continue;

<snip rest>

Sorry, I'm new to Groups. And thanks, the code works now!

[Keith Thompson]

hdsalbki <hd******@gmail .comwrites:

Thanks to everyone I was able to do this. apart from what vippstar has
sugested I also came with the below code to read the rest of line in
stdin. Is this okay or am I doing something wrong. Thanks for any
comments.

#include <stdio.h>

int main()
{
char a;
while(1)
{
printf("\nPleas e enter a character: ");
fflush(stdout);
scanf("%c",&a);
printf("You entered: %c\n",a);
fflush(stdout);
while(1)
{
scanf("%c",&a);
if(a=='\r' || a=='\n')
{
break;
}
}
}
return 0;
}

Why are you checking for '\r' as well as '\n'?

Even if your system happens to represent end-of-line with a CR-LF
pair, that will be translated to a single '\n' character on input (if
you're reading from a text stream, which you are here).

You're also not checking for end-of-file. (Hint: check for EOF, not
feof(stdin).)

I haven't read this whole thread yet, so I don't know whether anyone
else has suggested section 12 of the comp.lang.c FAQ, which covers
stdio. It's at <http://www.c-faq.com>.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

hdsalbki <hd******@gmail .comwrites:

santosh wrote:

#include <stdio.h>

int flushln(FILE *f) {
int ch;
while ((EOF != (ch = getc(f))) || (ch != '\n')) continue;

This needs to use the && logical operator to function as Chuck probably
intended it:

while ((EOF != (ch = getc(f))) && (ch != '\n')) continue;

<snip rest>

Sorry, I'm new to Groups. And thanks, the code works now!

It was Chuck's mistake, not yours; the code he posted actually had the
incorrect "||".

Not a huge deal either way; we all make mistakes. But it does
illustrate the fact that it's usually best to copy-and-paste source
code that's actually been successfully compiled and executed rather
than typing it on the fly.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

hdsalbki <hd******@gmail .comwrites:

I have below a small program to echo back what character the user
types. It's working OK but prints a extra prompt between every
character. what can I do? I'm new in C and my book is very
difficult...
thanks for any help.

#include<stdio. h>

main()
{
char a;
while(1)
{
printf("\nPleas e type a character: ");
scanf("%c",&a);
printf("\nYou typed: %c",a);
}
}

The program is actually doing just what you told it to do. In
response to the prompt, you typed, say, an 'x'. You didn't see
anything happen, so you pressed the return key (or enter, or however
it's labeled). You typed two characters, 'x' and return, and your
program reported each one.

The issue here is input buffering. (This is distinct from output
buffering, which I discussed in a separate followup.) Though your
scanf call asks for only one character, it doesn't complete until
you've entered a line of text. The new-line character is part of that
line.

More precisely, the scanf call *waits* for a full line of input, but
it only *consumes* a single character. Any other characters in the
line are left waiting in a buffer to be consumed by the next call to
scanf.

Try changing the second printf to:

printf("\nYou typed: '%c'",a);

so the output character is surrounded by single quotes. This will
show you more clearly that it's actually reading and printing each
character you type, including new-line.

Try typing just the return key, with nothing preceding it.

Try typing several characters, such as xyz<return>.

What you'd probably *like* to do is read a single character without
waiting for a complete line. C provides no guaranteed way to do that.
See question 19.1 of the comp.lang.c FAQ, <http://www.c-faq.com/>.

Or you can read a single character and then discard the remainder of
the input line. See the rest of this thread for various ways to do
that.

Text input in C is line-oriented. Usually the best way to handle it
is to use fgets() to read a full line at a time into a string, then
use other functions to process the string. You'll get to that later.
(Using fgets() introduces some issues for very long lines; that can
also wait for later.)

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

hdsalbki <hd******@gmail .comwrites:

I have below a small program to echo back what character the user
types. It's working OK but prints a extra prompt between every
character. what can I do? I'm new in C and my book is very
difficult...
thanks for any help.

#include<stdio. h>

main()
{
char a;
while(1)
{
printf("\nPleas e type a character: ");
scanf("%c",&a);
printf("\nYou typed: %c",a);
}
}

The program is actually doing just what you told it to do. In
response to the prompt, you typed, say, an 'x'. You didn't see
anything happen, so you pressed the return key (or enter, or however
it's labeled). You typed two characters, 'x' and return, and your
program reported each one.

The issue here is input buffering. (This is distinct from output
buffering, which I discussed in a separate followup.) Though your
scanf call asks for only one character, it doesn't complete until
you've entered a line of text. The new-line character is part of that
line.

More precisely, the scanf call *waits* for a full line of input, but
it only *consumes* a single character. Any other characters in the
line are left waiting in a buffer to be consumed by the next call to
scanf.

Try changing the second printf to:

printf("\nYou typed: '%c'",a);

so the output character is surrounded by single quotes. This will
show you more clearly that it's actually reading and printing each
character you type, including new-line.

Try typing just the return key, with nothing preceding it.

Try typing several characters, such as xyz<return>.

What you'd probably *like* to do is read a single character without
waiting for a complete line. C provides no guaranteed way to do that.
See question 19.1 of the comp.lang.c FAQ, <http://www.c-faq.com/>.

Or you can read a single character and then discard the remainder of
the input line. See the rest of this thread for various ways to do
that.

Text input in C is line-oriented. Usually the best way to handle it
is to use fgets() to read a full line at a time into a string, then
use other functions to process the string. You'll get to that later.
(Using fgets() introduces some issues for very long lines; that can
also wait for later.)

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

Keith Thompson <ks***@mib.orgw rites:

hdsalbki <hd******@gmail .comwrites:

>I have below a small program to echo back what character the user
types. It's working OK but prints a extra prompt between every
character. what can I do? I'm new in C and my book is very
difficult...

[snip]

The program is actually doing just what you told it to do.

[snip]

Apologies for the double post. I was insufficiently patient with my
news server.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Keith Thompson]

Keith Thompson <ks***@mib.orgw rites:

hdsalbki <hd******@gmail .comwrites:

>I have below a small program to echo back what character the user
types. It's working OK but prints a extra prompt between every
character. what can I do? I'm new in C and my book is very
difficult...

[snip]

The program is actually doing just what you told it to do.

[snip]

Apologies for the double post. I was insufficiently patient with my
news server.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Ben Bacarisse]

vi******@gmail. com writes:

On Jul 13, 4:56 pm, Ben Bacarisse <ben.use...@bsb .me.ukwrote:

<snip>

>The above fails because %*[\n] must match as least one character. The

Don't you mean %*[^\n]? Because that's what is used.

Yes, typo. Sorry. If I have the power to re-write standards I'd
allow a zero-length match for %[^...] since that is often what one
wants.

--
Ben.

[Ben Bacarisse]

"Joachim Schmitz" <no*********@sc hmitz-digital.dewrite s:
<snip>

>Why not
if(scanf("%c%*[^\r\n]%*c", &a) < 1) break;

I've got confused by corrections or corrections so maybe you mean
something else, but this code probably does not do what you expect.

As per my other post, the %*[^\r\n] must match at least one character.
If there is nothing but the \n, the whole scanf stops here having
failed (but it returns 1). The %*c does not get to read the newline.

--
Ben.