I apologize for the earlier post where I forgot the code.

I have included a portion of my code below. The problem that is driving me
crazy is that the first time thru the loop everything works fine but the
second time thru the loop the program just goes by the scanf statement. Could
someone please enlighten me as to what I am doing wrong.

Since my original post I have come across a function called _flushall() that
seems to flush out the stdin buffer. Why don't either of my two Introduction
to C Programming books talk about flushall() or problems with buffers?

int main(void)  
{
char y;
while (1)
        {
        printf("\nDo you want to input another string? (y/n)");
        scanf("%c",&y);
        if (y=='n') break;
    y='0';
    input();
    }
return 0;

Robert Sproat

| I apologize for the earlier post where I forgot the code.
|
| I have included a portion of my code below. The problem that is driving me
| crazy is that the first time thru the loop everything works fine but the
| second time thru the loop the program just goes by the scanf statement. Could
| someone please enlighten me as to what I am doing wrong.

| Since my original post I have come across a function called _flushall() that
| seems to flush out the stdin buffer. Why don't either of my two Introduction
| to C Programming books talk about flushall() or problems with buffers?

That could be because flushall() and _flushall() are not part of C.  They
may be a specialized extension of a given implementation.  It may be to
cover up for weaknesses in their library or weak programmers.  But you
don't need them.

| int main(void)  
| {
| char y;
| while (1)
|       {
|       printf("\nDo you want to input another string? (y/n)");
|       scanf("%c",&y);
|       if (y=='n') break;
|     y='0';
|     input();
|     }
| return 0;
| }/* End Main */

Did you have to type return to get the first input to work?  Your
operating system may have built a complete line of input first,
which in your case was one letter, but the input actually has both
that letter and a newline in the buffer.  The scanf gets the letter
the first time through, but the newline is still there for the next
time.

You might try:

    int n;
    ...
    scanf("%c\n%n",&y,&n);

This will set "n" with the number of characters read from the input
stream, being 1 if the "\n" didn't match (i.e. more than one character
was typed before the end of line) and being 2 if the "\n" did match.

Alternatively, use getc() or getchar() to read from input until you
find the "\n" (or EOF) then it should be safe to take another line
of input (if not EOF ... but depending on OS, it may be anyway).

If you want "hot key" input, you will have to apply operating system
tricks to get that to work, and of course that's OS dependent and you
need to pick the correct OS related newsgroup.

--
Phil Howard KA9WGN   +-------------------------------------------------------+
Linux Consultant     |  "Adolf Hitler was OK at the beginning, but then he   |
Milepost Services    |  went too far".   --Marge Schott    --World History   |

: >I apologize for the earlier post where I forgot the code.
: >
: >I have included a portion of my code below. The problem that is driving me
: >crazy is that the first time thru the loop everything works fine but the
: >second time thru the loop the program just goes by the scanf statement. Could
: >someone please enlighten me as to what I am doing wrong.
: >
: >Since my original post I have come across a function called _flushall() that
: >seems to flush out the stdin buffer. Why don't either of my two Introduction
: >to C Programming books talk about flushall() or problems with buffers?
: >
: >int main(void)  
: >{
: >char y;
: >while (1)
: >  {
: >  printf("\nDo you want to input another string? (y/n)");

: You can't use printf() without a prototype.

You can use printf without a prototype because it returns an int, and
because compilers still must support K&R C. Of course it is better to use
prototypes and include stdio.h.

: >  scanf("%c",&y);
As above.

: You can't use scanf() without a prototype.  Also, this is
: very poor practice.  Using fgets() and sscanf() are much
: better.

You can use scanf() if you know how. I would not use it myself because I
just made a mistake when I was telling Robert how to use it myself.

: >  if (y=='n') break; : >    y='0';
: >    input();

: What is this?  There is no input() in the standard library.
How do you know input() is not a function written by Robert?
The only real bad thing here is that he would want to use a non-standard
function like flushall(), which incidentally you did not notice. :-)

: >    }
: >return 0;
: >}/* End Main */

: In the future, try compiling your code.

--
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Canada

No beast so fierce but knows some touch of pity
But I know none, And therefore am no beast
              Richard III., William Shakespeare
vim : the best editor
ftp.fu-berlin.de/misc/editors/vim/beta-test
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"But it works on any compiler I've used" is not a valid argument in this
newsgroup.

Dan
--
Dan Pop
CERN, CN Division

Mail:  CERN - PPE, Bat. 31 R-004, CH-1211 Geneve 23, Switzerland

: GE: : You can't use printf() without a prototype.
: GE:
: GE: You can use printf without a prototype because it returns an int, and
: GE: because compilers still must support K&R C. Of course it is better to use
: GE: prototypes and include stdio.h.
: GE:

: The standard specifically states that varargs functions *must* have
: prototypes in scope when they are called. So, printf needs a
: prototype.

: By the way, not all valid K&R I C is valid ANSI C. There were a few
: changes, and one has to adhere to them. For example, the following is
: perfectly valid K&R I C, but try compiling it on a truly ANSI C
: compiler!

: #include <stdio.h>
: int main() { printf("/Is everything okay??/"); return 0; }

You would expect this to generate something like unterminated string
constant. However, when I compile with gcc without the -ansi switch it
works fine. :-)

: Cheers
: Tanmoy
: --

: Tanmoy Bhattacharya O:T-8(MS B285)LANL,NM87545 H:#9,3000,Trinity Drive,NM87544
: Others see <gopher://yaleinfo.yale.edu:7700/00/Internet-People/internet-mail>,
: <http://alpha.acast.nova.edu/cgi-bin/inmgq.pl>or<ftp://csd4.csd.uwm.edu/pub/
: internetwork-mail-guide>. -- <http://nqcd.lanl.gov/people/tanmoy/tanmoy.html>
: fax: 1 (505) 665 3003   voice: 1 (505) 665 4733    [ Home: 1 (505) 662 5596 ]

--
---------------------------------------------------------------------


Canada

No beast so fierce but knows some touch of pity
But I know none, And therefore am no beast
              Richard III., William Shakespeare
vim : the best editor
ftp.fu-berlin.de/misc/editors/vim/beta-test
---------------------------------------------------------------------

: >
: >: You can't use printf() without a prototype.
: >
: >You can use printf without a prototype because it returns an int, and
: >because compilers still must support K&R C.

: Please don't post nonsense.  You can't use _any_ variadic function
: without a prototype in scope.  printf is no exception.  Compilers are
: NOT required to support K&R C code which is invoking undefined behaviour
: according to the ANSI C standard.

: "But it works on any compiler I've used" is not a valid argument in this
: newsgroup.

I won't use that argument. In fact what I am about to say is not fact but
my personal view on this.
The prototype for printf in this case, without a prototype is
int printf()
That means that nothing should be assumed about the arguments to printf
by the compiler. Now this should work unless you do something really
boneheaded on a system where ints are 16 bit and longs are 32 bit and
pointers are 32 bit.
eg. printf("%p",0); could cause problems.
or printf("%ld",10); could also cause problems, I agree on that.
But scanf is variadic and should work unless you do not pass the address
of a variable which is wrong no matter how you look at it.
Of course I am not saying that you should not prototype or that you do
not include the appropriate headers. Also you would use varargs.h for
variadic functions in K&R C. Handling varargs the way they are now was
actually borrowed from C++. It was handled quite differently in K&R C.

: Dan
: --
: Dan Pop
: CERN, CN Division

: Mail:  CERN - PPE, Bat. 31 R-004, CH-1211 Geneve 23, Switzerland

--
---------------------------------------------------------------------


Canada

No beast so fierce but knows some touch of pity
But I know none, And therefore am no beast
              Richard III., William Shakespeare
vim : the best editor
ftp.fu-berlin.de/misc/editors/vim/beta-test
---------------------------------------------------------------------

Dan
--
Dan Pop
CERN, CN Division

Mail:  CERN - PPE, Bat. 31 R-004, CH-1211 Geneve 23, Switzerland

Not every declaration is a prototype.

Dan
--
Dan Pop
CERN, CN Division

Mail:  CERN - PPE, Bat. 31 R-004, CH-1211 Geneve 23, Switzerland

That is now (at least) the third time the point has been made in this
thread alone. Surely that is proof enough to bring closure to any
ill-advised debating otherwise.

--
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      Linux for fun, M$ for $$$...and the NFL for what really counts!
=============================================================================

Thank you everyone for your help. I just want to make it clear that I just
gave you the portion of my code where I thought the problem was (to keep the
posting small). In original code I did include stdio.h and input() was one of
my own functions. I have learned a lesson, next time I will post everything
relevant.

int printf(const char *, ...);

The standard states that calling a function defined with a prototype
containing ... (such as the definition of printf in the standard) without
an appropriate prototype in scope at the point of call, results in undefined
behaviour.

The rules in the C language are clearly intended to permit a C compiler to
assime that all functions have fixed arguments except where it sees an
explicit prototype with a ...

--
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