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);
}
}

[hdsalbki]

vipps...@gmail. com wrote:

On Jul 13, 2:58 pm, hdsalbki <hdsal...@gmail .comwrote:

okay I followed everyone ideas and wrote another version of the
program. Now there is no extra prompt but the code misses every
alternate character I type. Can anyone help? Thanks a lot, never
imagined such a simple thing will be so difficult!

#include <stdio.h>

int main()

I told you int main(void), not int main().

{
char a;
while(1)
{
printf("\nPleas e enter a character: ");
fflush(stdout);
if(scanf("%c%*[^\n]%*c",&a) < 1)
{
break;
}
printf("You typed: %c\n",a);
fflush(stdout);

That fflush() is not necessary, as you write a newline to stdout. That
will flush the stream, since it's line buffered. (_IOLBF in setvbuf)

}
return(0);

return is not a function, it's a keyword. the parenthesis are not
needed, just like sizeof. (do not confuse it with sizeof (cast), it's
the cast that requires parenthesis, not sizeof)

}

o/p is:

dodo@sapphire:~/src/c/ws$ ./echo_char1

Please enter a character: a
You typed: a

Please enter a character: b
You typed:

Please enter a character: c
You typed: c

Please enter a character: d
You typed:

Please enter a character:

Yeah, it seems that it does not actually read the newline with %*c.
I can't tell you why this happends, perhaps someone else can explain.
Remove the %*c and add a getchar(), and it should work:

if(scanf("%c%^[\n]", &a) < 1) break;
(void)getchar() ;

Note that the cast to (void) is not necessary; I only added it to make
more clear that you don't need to check the return value of getchar.

Thank you vippstar. I will use int main(void) from now on and use
fflush(stdout) only when there is no \n after output. Okay I confused
the return keyword with if() and while() which need brackets. And your
change works! looking at info for getchar() it says it is for reading
single character...may be I should've done it with getchar()...tha nks
again.

[vippstar]

On Jul 13, 4:35 pm, hdsalbki <hdsal...@gmail .comwrote:

hdsalbki wrote:

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.

<snip wrong code>

>
Sorry some errors were there. Corrected code follows:

<snip wrong code>

Both are wrong. My code is correct, compare both codes and try to
understand why yours is not correct.
Do not continue this discussion if you're not a troll. Just reread the
discussion and use a reference for all the functions you will use.

[Joachim Schmitz]

hdsalbki wrote:

hdsalbki wrote:

>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;
}

Sorry some errors were there. Corrected code follows:

#include <stdio.h>

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

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

}
}
return 0;
}

Bye, Jojo

[Joachim Schmitz]

Joachim Schmitz wrote:

hdsalbki wrote:

>hdsalbki wrote:

>>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;
}

Sorry some errors were there. Corrected code follows:

#include <stdio.h>

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

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

Oops. Should be in the outer while, the inner while deleted entirely.

>

> }
}
return 0;
}

Bye, Jojo

[Ben Bacarisse]

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

okay I followed everyone ideas and wrote another version of the
program. Now there is no extra prompt but the code misses every
alternate character I type. Can anyone help? Thanks a lot, never
imagined such a simple thing will be so difficult!

#include <stdio.h>

int main()
{
char a;
while(1)
{
printf("\nPleas e enter a character: ");
fflush(stdout);
if(scanf("%c%*[^\n]%*c",&a) < 1)
{
break;
}
printf("You typed: %c\n",a);
fflush(stdout);
}
return(0);
}

You may have thought you picked an easy example, but this is very hard
using scanf. I simply would not try it. All the suggestions so far
are wrong in subtle ways.

The above fails because %*[\n] must match as least one character. The
match fails at that point (if there is nothing before the \n) and the
whole scanf return without doing the %*c (read without assigning a
char).

I'd use a getc call and a look to read and discard any remaining
characters up to the newline.

--
Ben.

[vippstar]

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

hdsalbki <hdsal...@gmail .comwrites:

okay I followed everyone ideas and wrote another version of the
program. Now there is no extra prompt but the code misses every
alternate character I type. Can anyone help? Thanks a lot, never
imagined such a simple thing will be so difficult!

<snip code>

>
You may have thought you picked an easy example, but this is very hard
using scanf. I simply would not try it. All the suggestions so far
are wrong in subtle ways.

My code along with the getchar() fix is fine.

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

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

match fails at that point (if there is nothing before the \n) and the
whole scanf return without doing the %*c (read without assigning a
char).

Ah, so that's why that happends. Thanks.
<snip>

[hdsalbki]

Ben Bacarisse wrote:

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

okay I followed everyone ideas and wrote another version of the
program. Now there is no extra prompt but the code misses every
alternate character I type. Can anyone help? Thanks a lot, never
imagined such a simple thing will be so difficult!

#include <stdio.h>

int main()
{
char a;
while(1)
{
printf("\nPleas e enter a character: ");
fflush(stdout);
if(scanf("%c%*[^\n]%*c",&a) < 1)
{
break;
}
printf("You typed: %c\n",a);
fflush(stdout);
}
return(0);
}

You may have thought you picked an easy example, but this is very hard
using scanf. I simply would not try it. All the suggestions so far
are wrong in subtle ways.

The above fails because %*[\n] must match as least one character. The
match fails at that point (if there is nothing before the \n) and the
whole scanf return without doing the %*c (read without assigning a
char).

I'd use a getc call and a look to read and discard any remaining
characters up to the newline.

--
Ben.

Thanks Ben. I see now that getc() is suited for this. I used scanf
because I have not yet arrived in my book where getchar()/getc() is
used.

[hdsalbki]

vipps...@gmail. com wrote:

On Jul 13, 4:35 pm, hdsalbki <hdsal...@gmail .comwrote:

hdsalbki wrote:

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.

<snip wrong code>

Sorry some errors were there. Corrected code follows:

<snip wrong code>

Both are wrong. My code is correct, compare both codes and try to
understand why yours is not correct.
Do not continue this discussion if you're not a troll. Just reread the
discussion and use a reference for all the functions you will use.

Sorry if I offended you. I thought my version was correct because it
behaved just like your code. Anyway thanks for your help...i'm moving
on to the next exercise!

[CBFalconer]

hdsalbki wrote:

vipps...@gmail. com wrote:

.... snip ...

>

>Note that the cast to (void) is not necessary; I only added it
to make more clear that you don't need to check the return value
of getchar.

Thank you vippstar. I will use int main(void) from now on and use
fflush(stdout) only when there is no \n after output. Okay I
confused the return keyword with if() and while() which need
brackets. And your change works! looking at info for getchar() it
says it is for reading single character...may be I should've done
it with getchar()...tha nks again.

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.

[hdsalbki]

CBFalconer wrote:

hdsalbki wrote:

vipps...@gmail. com wrote:

... snip ...

Note that the cast to (void) is not necessary; I only added it
to make more clear that you don't need to check the return value
of getchar.

Thank you vippstar. I will use int main(void) from now on and use
fflush(stdout) only when there is no \n after output. Okay I
confused the return keyword with if() and while() which need
brackets. And your change works! looking at info for getchar() it
says it is for reading single character...may be I should've done
it with getchar()...tha nks again.

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.

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;
return ch;
} /* flushln */

int main(void)
{
char a;
while(1)
{
printf("Please enter a character: ");
fflush(stdout);
scanf("%c",&a);
printf("You typed: %c\n",a);
flushln(stdin);
}
return 0;
}

o/p is:

dodo@sapphire:~/src/c/ws$ ./echo_char3
Please enter a character: x
You typed: x

Program hangs here. whatever I type the prompt is not printing again.