Help. Where is my error?

[Red Dragon]

I am self study C student. I got stuck in the program below on quadratic equation and will be most grateful if someone could help me to unravel the mystery.
Why does the computer refuse to execute my scanf ("%c",&q);
On input 3 4 1 (for a,b and c) I had real roots OK
On input 1 8 16 I had same real roots OK.

However on 4 2 5, (for imaginary roots ) the computer cannot see the scanf ("%c",&q); statement. It just jumps over it.
How can I make the computer not to ignore this statement? I am on Visual C++ platform.
Thanks
Khoon.

/* Roots of a Quadratic Equation.
12.10.05 */

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main (void)

{
int a; int b; int c; float x1; float x2; int E; int E1; float R; float I;float S;
char p; char q; char y;

printf ("Please key in the value of constant a,b and c for finding the roots of quadratic");
printf ("equation ax%c+bx+c=0 :",253);
scanf ("%d%d%d", &a,&b,&c);

E =(b*b)-(4*a*c);

if ( E > 0)
{
x1 = (float)(-b+sqrt(E))/(2*a);
x2 = (float)(-b-sqrt(E))/(2*a);

printf ("\nYour quadratic equation has two distinct real roots: x1=%1.6f ,x2=%1.6f",x1,x 2);
}

else if (E == 0)
{

x1 = (float)(-b+sqrt(E))/(2*a);

printf ("\nYour quadratic equation has two same: x1=x2=%1.6f\n", x1);
}

else
{

p = 'y';

printf ("Your quadratic equation has two distinct imaginary roots. Do you want to know\n");
printf ("the values of the imaginary roots (Y/N)?");

scanf ("%c",&q);

printf ("\nq = %c\n",q);/* Test statement*/

if (p==q)
printf ("OK I will show your the imaginary roots tomorrow.\n");

else
printf ("Good bye\n");

return 0;

}
}

[Red Dragon]

"Old Wolf" <ol*****@inspir e.net.nz> wrote in message news:11******** *************@g 43g2000cwa.goog legroups.com...

Red Dragon wrote:

I am self study C student. I got stuck in the program below on
quadratic equation and will be most grateful if someone could help
me to unravel the mystery.
Why does the computer refuse to execute my scanf ("%c",&q);
You posted this program a couple of weeks ago and got lots
of advice. But you seem to have ignored most of this advice.
Here it is again:

1) x1, x2, R, I, S should be "double", not "float"
Thank you for your help and I very much appreciate it. I am doing the exercise in my book and since precision is not an issue here, I have put it on lower priority. But as you rightly point out. I will adopt it as I will be dealing with big figures.
2) You MUST check the return value of scanf() and take appropriate
action if it is wrong
I was doing this. That is why I call it "test statement".
3) Don't do any casting (eg. in the line "x1 = (float)FOO")
4) The "return 0" goes AFTER the "else" block, not inside it.

Thanks
Khoon.

[pete]

Red Dragon wrote:

"Old Wolf" <ol*****@inspir e.net.nz> wrote in message
news:11******** *************@g 43g2000cwa.goog legroups.com...

You posted this program a couple of weeks ago and got lots
of advice. But you seem to have ignored most of this advice.
Here it is again:
1) x1, x2, R, I, S should be "double", not "float"

Thank you for your help and I very much appreciate it. I am doing the
exercise in my book and since precision is not an issue here, I have
put it on lower priority. But as you rightly point out. I will
adopt it as I will be dealing with big figures.

What are you waiting for?
Unless you have an array of them,
small arithmetic types like float, short, and char,
which often get quietly converted to larger types,
are very probably a poor choice.

2) You MUST check the return value of scanf() and take appropriate
action if it is wrong

I was doing this. That is why I call it "test statement".

int rc;

rc = scanf("%c", &q);
printf("scanf returned a value of %d\n", rc); /* Test statement*/

--
pete

[Red Dragon]

I have got all the results by using 2 scanf(). One is a dummy. I am also using double instead of float.
Below are the results. I am copying and paste from Notepad, so I hope I dont have HTML problem.
Regards,
Khoon
/* Roots of a Quadratic Equation.
12.10.05 */

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main (void)

{
int a; int b; int c; double x1; double x2; double E; int E1; double R; double I;double S;
char q; char Y;

printf ("Please key in the value of constant a,b and c for finding the roots of quadratic");
printf ("equation ax%c+bx+c=0 :",253);
scanf ("%d%d%d", &a,&b,&c);

E =(b*b)-(4*a*c);

if ( E > 0)
{
x1 = (float)(-b+sqrt(E))/(2*a);
x2 = (float)(-b-sqrt(E))/(2*a);

printf ("\nYour quadratic equation has two distinct real roots: x1=%1.6f ,x2=%1.6f",x1,x 2);
}

else if (E == 0)
{

x1 = (float)(-b+sqrt(E))/(2*a);

printf ("\nYour quadratic equation has two same: x1=x2=%1.6f\n", x1);
}

else

{

printf ("Your quadratic equation has two distinct imaginary roots. Do you want to know");
printf ("\nthe values of the imaginary roots (Y/N)?");

scanf ("%c",&q); /* Dummy. The computer jumps this scanf() */

scanf ("%c",&q);

printf ("q = %c\n",q); /* Test statement*/

if ('Y'==q)
{
R = (float)-b/(2*a);
S=abs(E);
S=sqrt(S);
I = S/(2*a);
printf ("\nThe imaginary roots are:\n");
printf (" x1=%1.6f + %1.6fi , x2=%1.6f - %1.6fi\n",R,I,R ,I);
}
else
printf ("Thank you for using this computer\n");
}
return 0;
}

/* RESULT
Please key in the value of constant a,b and c for finding the roots of quadratic
equation ax²+bx+c=0 :3 4 1

Your quadratic equation has two distinct real roots: x1=-0.333333 ,x2=-1.000000
Press any key to continue */

/*Please key in the value of constant a,b and c for finding the roots of quadratic
equation ax²+bx+c=0 :1 8 16

Your quadratic equation has two same: x1=x2=-4.000000
Press any key to continue */

/*Please key in the value of constant a,b and c for finding the roots of quadrati
equation ax²+bx+c=0 :4 2 5
Your quadratic equation has two distinct imaginary roots. Do you want to know
the values of the imaginary roots (Y/N)?Y
q = Y

The imaginary roots are:
x1=-0.250000 + 1.089725i , x2=-0.250000 - 1.089725i
Press any key to continue*/

/*Please key in the value of constant a,b and c for finding the roots of quadratic
equation ax²+bx+c=0 :4 2 5
Your quadratic equation has two distinct imaginary roots. Do you want to know
the values of the imaginary roots (Y/N)?N
q = N
Thank you for using this computer
Press any key to continue*/

[Christopher Benson-Manica]

Red Dragon <ts*****@stream yx.com> wrote:

1. The dont understand the HTML thing. I did not use HTML. What I did was
to copy from my Visual C++ platform and paste on the Outlook Express
screen. With Mr. Skarmander's instruction, I have set the radio button to
"Plain Text" on the Send Tab and thought the problem is solved. I have
actually sent a copy of the outgoing mail to myself and I dont see any HTML
thing on my screen on the returned copy. So I dont know what else to do.

Well, whatever you've done worked for this post, at least - we'll let
you know if the problem occurs again.

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[Red Dragon]

> What are you waiting for?

Unless you have an array of them,
small arithmetic types like float, short, and char,
which often get quietly converted to larger types,
are very probably a poor choice.

> 2) You MUST check the return value of scanf() and take appropriate
> action if it is wrong

I was doing this. That is why I call it "test statement".

int rc;

rc = scanf("%c", &q);
printf("scanf returned a value of %d\n", rc); /* Test statement*/

--
pete

Hi Pete,
I have already been using double.
As for the test, I find that for all inputs of q, "rc" registers 1.
So what is it suppose to test? What does it suppose to show?
I cannot see how to make use of it because everything I do, "scanf returned
a value of 1".
Thanks
Khoon.

[Red Dragon]

> Well, whatever you've done worked for this post, at least - we'll let

you know if the problem occurs again.

This post of course works because I dont have any C- program loaded on it.
I have yet to see when I paste a C-program on it, what will be its
outcome.

Now I paste from Notepad and not directly from Visual C++ platform.
Please let me know if the problem occurs again.
Thanks
Khoon.

[Tim Rentsch]

"Red Dragon" <ts*****@stream yx.com> writes:

You have been repeatedly asked not to post using HTML.
Please stop it. This is usenet, not some mailing list or forum.

[snip]
Mr. Michael,
I wish to apologize for causing you problem. I did not know I was causing a
problem.
1. The dont understand the HTML thing. I did not use HTML. What I did was
to copy from my Visual C++ platform and paste on the Outlook Express
screen. With Mr. Skarmander's instruction, I have set the radio button to
"Plain Text" on the Send Tab and thought the problem is solved. I have
actually sent a copy of the outgoing mail to myself and I dont see any HTML
thing on my screen on the returned copy. So I dont know what else to do.
I suggest trying a mail tool other than Microsoft Outlook Express.
Microsoft mail software is well known for behaving in ways that
can cause problems like this.

Probably what you're using to read news hides the HTML-ness
of what you're posting. FYI, here is what gets transmitted
(each line has '>->-> ' at the beginning). I expect you can
see why other people don't like reading postings like this.

->-> From: "Red Dragon" <ts*****@stream yx.com>
->-> Subject: Re: Help. Where is my error?
->-> Newsgroups: comp.lang.c
->-> Date: Tue, 18 Oct 2005 20:43:20 +0800
->-> Organization: TMnet Malaysia References: <43********@new s.tm.net.my> <11************ *********@g43g2 000cwa.googlegr oups.com> <43**********@n ews.tm.net.my>
Lines: 281
MIME-Version: 1.0
Content-Type: multipart/alternative;
boundary="----=_NextPart_000_ 000E_01C5D424.9 4159240"
X-Newsreader: Microsoft Outlook Express 6.00.2900.2180
X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2180
NNTP-Posting-Host: 218.111.62.120
X-Original-NNTP-Posting-Host: 218.111.62.120
Message-ID: <43********@new s.tm.net.my>
X-Trace: news.tm.net.my 1129639409 218.111.62.120 (18 Oct 2005 20:43:29 +0800)
Path: nntp-server.caltech. edu!hammer.uore gon.edu!news.gl orb.com!news-feed01.roc.ny.f rontiernet.net! nntp.frontierne t.net!uunet!spo ol.news.uu.net! ash.uu.net!news 1.tm.net.my!not-for-mail->->
->-> This is a multi-part message in MIME format.
->->
->-> ------=_NextPart_000_ 000E_01C5D424.9 4159240
->-> Content-Type: text/plain;
->-> charset="iso-8859-1"
->-> Content-Transfer-Encoding: quoted-printable
->->
->-> I have got all the results by using 2 scanf(). One is a dummy. I am =
->-> also using double instead of float.=20
->-> Below are the results. I am copying and paste from Notepad, so I hope =
->-> I dont have HTML problem.=20
->-> Regards,
->-> Khoon
->->
->->
->-> /* Roots of a Quadratic Equation.
->-> 12.10.05 */
->->
->-> #include <stdio.h>
->-> #include <stdlib.h>
->-> #include <math.h>
->->
->-> int main (void)
->->
->-> {
->-> int a; int b; int c; double x1; double x2; double E; int E1; double R; =
->-> double I;double S;
->-> char q; char Y;
->->
->-> printf ("Please key in the value of constant a,b and c for finding the =
->-> roots of quadratic");
->-> printf ("equation ax%c+bx+c=3D0 :",253);
->-> scanf ("%d%d%d", &a,&b,&c);
->-> =20
->-> E =3D(b*b)-(4*a*c);
->-> =20
->-> if ( E > 0)=20
->-> { =20
->-> x1 =3D (float)(-b+sqrt(E))/(2*a);
->-> x2 =3D (float)(-b-sqrt(E))/(2*a);
->->
->-> printf ("\nYour quadratic equation has two distinct real roots: =
->-> x1=3D%1.6f ,x2=3D%1.6f",x1 ,x2);
->-> }
->-> =20
->-> else if (E =3D=3D 0)=20
->-> {
->->
->-> x1 =3D (float)(-b+sqrt(E))/(2*a);
->-> =20
->-> printf ("\nYour quadratic equation has two same: =
->-> x1=3Dx2=3D%1.6f \n",x1);
->-> }
->-> =20
->-> else=20
->->
->-> {
->->
->-> printf ("Your quadratic equation has two distinct imaginary roots. Do =
->-> you want to know");
->-> printf ("\nthe values of the imaginary roots (Y/N)?");
->-> =20
->-> scanf ("%c",&q); /* Dummy. The computer jumps this =
->-> scanf() */
->->
->-> scanf ("%c",&q);
->->
->-> printf ("q =3D %c\n",q); /* Test statement*/
->->
->-> if ('Y'=3D=3Dq)
->-> {
->-> R =3D (float)-b/(2*a);
->-> S=3Dabs(E);
->-> S=3Dsqrt(S);
->-> I =3D S/(2*a);
->-> printf ("\nThe imaginary roots are:\n");
->-> printf (" x1=3D%1.6f + %1.6fi , x2=3D%1.6f - %1.6fi\n",R,I,R ,I);
->-> }
->-> else
->-> printf ("Thank you for using this computer\n");
->-> }
->-> return 0;
->-> }
->->
->-> /* RESULT
->-> Please key in the value of constant a,b and c for finding the roots of =
->-> quadratic
->-> equation ax=B2+bx+c=3D0 :3 4 1
->->
->-> Your quadratic equation has two distinct real roots: x1=3D-0.333333 =
->-> ,x2=3D-1.000000
->-> Press any key to continue */
->->
->-> /*Please key in the value of constant a,b and c for finding the roots of =
->-> quadratic
->-> equation ax=B2+bx+c=3D0 :1 8 16
->->
->-> Your quadratic equation has two same: x1=3Dx2=3D-4.000000
->-> Press any key to continue */
->->
->-> /*Please key in the value of constant a,b and c for finding the roots of =
->-> quadrati
->-> equation ax=B2+bx+c=3D0 :4 2 5
->-> Your quadratic equation has two distinct imaginary roots. Do you want =
->-> to know
->-> the values of the imaginary roots (Y/N)?Y
->-> q =3D Y
->->
->-> The imaginary roots are:
->-> x1=3D-0.250000 + 1.089725i , x2=3D-0.250000 - 1.089725i
->-> Press any key to continue*/
->->
->-> /*Please key in the value of constant a,b and c for finding the roots of =
->-> quadratic
->-> equation ax=B2+bx+c=3D0 :4 2 5
->-> Your quadratic equation has two distinct imaginary roots. Do you want =
->-> to know
->-> the values of the imaginary roots (Y/N)?N
->-> q =3D N
->-> Thank you for using this computer
->-> Press any key to continue*/
->-> =20
->->
->-> ------=_NextPart_000_ 000E_01C5D424.9 4159240
->-> Content-Type: text/html;
->-> charset="iso-8859-1"
->-> Content-Transfer-Encoding: quoted-printable
->->
->-> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
->-> <HTML><HEAD>
->-> <META http-equiv=3DContent-Type content=3D"text/html; =
->-> charset=3Diso-8859-1">
->-> <META content=3D"MSHT ML 6.00.2900.2769" name=3DGENERATO R>
->-> <STYLE></STYLE>
->-> </HEAD>
->-> <BODY bgColor=3D#ffff ff>
->-> <DIV><FONT face=3DArial size=3D2>I have got all the results by using 2=20
->-> scanf().&nbsp; One is a dummy. &nbsp;I am also using double instead of =
->-> float.=20
->-> </FONT></DIV>
->-> <DIV><FONT face=3DArial size=3D2>Below are the results.&nbsp; I am =
->-> copying&nbsp; and=20
->-> paste from Notepad, so I hope I dont have HTML problem. </FONT></DIV>
->-> <DIV><FONT face=3DArial size=3D2>Regard s,</FONT></DIV>
->-> <DIV><FONT face=3DArial size=3D2>Khoon</FONT></DIV>
->-> <DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
->-> <DIV>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial size=3D2>&nbsp; <FONT =
->-> color=3D#000080 ><STRONG>/*&nbsp;&nbsp;=2 0
->-> Roots of a Quadratic Equation.<BR>&n bsp;&nbsp;&nbsp ;&nbsp;&nbsp; =
->-> 12.10.05&nbsp;= 20
->-> */</STRONG></FONT></FONT></DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 =
->-> size=3D2><STRON G></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>#include=20
->-> &lt;stdio.h&gt; <BR>#include &lt;stdlib.h&gt ;<BR>#include=2 0
->-> &lt;math.h&g t;</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>int main=20
->-> (void)</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>{<BR>&nbsp;in t =
->-> a; int b; int=20
->-> c; double x1; double x2; double E; int E1; double R; double I;double=20
->-> S;<BR>&nbsp;&nb sp;&nbsp; char q; char Y;</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>&nbsp;printf =
->-> ("Please key in=20
->-> the value of constant a,b and c for finding the roots of=20
->-> quadratic");<BR >&nbsp;&nbsp;&n bsp; printf ("equation ax%c+bx+c=3D0&n bsp; =
->->
->-> :",253);<BR>&nb sp;scanf ("%d%d%d", =
->-> &amp;a,&amp;b,& amp;c);<BR>&nbs p;&nbsp;&nbsp;= 20
->-> <BR>&nbsp;&nbsp ;&nbsp; E&nbsp; =
->-> =3D(b*b)-(4*a*c);<BR>&nb sp;<BR>&nbsp;&n bsp; if ( E=20
->-> &gt; 0) <BR>&nbsp;&nbsp ; {&nbsp;&nbsp;&n bsp;&nbsp; =
->-> <BR>&nbsp;&nbsp ;&nbsp; x1 =3D=20
->-> (float)(-b+sqrt(E))/(2*a);<BR>&nbsp ;&nbsp;&nbsp; x2 =3D=20
->-> (float)(-b-sqrt(E))/(2*a);</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 =
->-> size=3D2><STRON G>&nbsp;&nbsp;p rintf ("\nYour=20
->-> quadratic equation has two distinct real roots: x1=3D%1.6f=20
->-> ,x2=3D%1.6f",x1 ,x2);<BR>&nbsp; &nbsp; }<BR>&nbsp;&nbs p; <BR>&nbsp;&nbsp ; =
->-> else if (E=20
->-> =3D=3D 0) <BR>&nbsp;&nbsp ; {</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 =20
->-> size=3D2><STRON G>&nbsp;&nbsp;& nbsp;&nbsp;&nbs p; x1 =3D=20
->-> (float)(-b+sqrt(E))/(2*a);<BR>&nbsp ;&nbsp;&nbsp;&n bsp;&nbsp;&nbsp ;=20
->-> <BR>&nbsp;&nbsp ; printf ("\nYour quadratic equation has two same:=20
->-> x1=3Dx2=3D%1.6f \n",x1);<BR>&nb sp;&nbsp; }<BR>&nbsp; <BR>&nbsp;&nbsp ; =
->-> else=20
->-> </STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>&nbsp;&nbsp;= 20
->-> {</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>&nbsp; printf =
->-> ("Your=20
->-> quadratic equation has two distinct imaginary roots.&nbsp; Do you want =
->-> to=20
->-> know");<BR>&nbs p; printf ("\nthe values of the imaginary roots=20
->-> (Y/N)?");<BR>&nbsp ;<BR>&nbsp; scanf=20
->-> ("%c",&amp;q );</STRONG>&nbsp;&n bsp;&nbsp;&nbsp ;&nbsp;&nbsp;&n bsp;&nbsp;&n=
->-> bsp;&nbsp;&nbsp ;&nbsp;=20
->-> &nbsp;<FONT color=3D#008080 ><STRONG>/*&nbsp;Dummy. &nbsp;The computer =
->-> jumps this=20
->-> scanf() */</STRONG></FONT></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>&nbsp; scanf=20
->-> ("%c",&amp;q );</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>&nbsp; printf =
->-> ("q =3D=20
->-> %c\n",q<FONT color=3D#008080 >)<FONT=20
->-> color=3D#000080 >;&nbsp;</FONT>&nbsp;&nbs p;&nbsp;&nbsp; &nbsp;/* Test=20
->-> statement*/</FONT></STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>&nbsp; if =
->-> ('Y'=3D=3Dq)<BR >&nbsp;=20
->-> {<BR>&nbsp;&nbs p;&nbsp;&nbsp;& nbsp; R =3D=20
->-> (float)-b/(2*a);<BR>&nbsp ;&nbsp;&nbsp;&n bsp;&nbsp; =
->-> S=3Dabs(E);<BR> &nbsp;&nbsp; =20
->-> S=3Dsqrt(S);<BR >&nbsp;&nbsp; I =3D =
->-> S/(2*a);<BR>&nbsp ;&nbsp;&nbsp;&n bsp;&nbsp; printf=20
->-> ("\nThe imaginary roots are:\n");<BR>&n bsp;&nbsp;&nbsp ;&nbsp;&nbsp; =
->-> printf ("=20
->-> x1=3D%1.6f + %1.6fi , x2=3D%1.6f - %1.6fi\n",R,I,R ,I);<BR>&nbsp; =
->-> }<BR>&nbsp;=20
->-> else<BR>&nbsp;& nbsp; printf ("Thank you for using this=20
->-> computer\n");<B R>}<BR>&nbsp; return 0;<BR>}</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>/* =
->-> RESULT<BR>Pleas e key in=20
->-> the value of constant a,b and c for finding the roots of =
->-> quadratic<BR>eq uation=20
->-> ax=B2+bx+c=3D0& nbsp; :3 4 1</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>Your quadratic =
->-> equation has=20
->-> two distinct real roots: x1=3D-0.333333 ,x2=3D-1.000000<BR>Pre ss any key =
->-> to continue=20
->-> */</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>/*Please key in =
->-> the value of=20
->-> constant a,b and c for finding the roots of quadratic<BR>eq uation=20
->-> ax=B2+bx+c=3D0& nbsp; :1 8 16</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>Your quadratic =
->-> equation has=20
->-> two same: x1=3Dx2=3D-4.000000<BR>Pre ss any key to continue =
->-> */</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>/*Please key in =
->-> the value of=20
->-> constant a,b and c for finding the roots of quadrati<BR>equ ation=20
->-> ax=B2+bx+c=3D0& nbsp; :4 2 5<BR>Your quadratic equation has two distinct =
->-> imaginary=20
->-> roots.&nbsp; Do you want to know<BR>the values of the imaginary roots=20
->-> (Y/N)?Y<BR>q =3D Y</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>The imaginary =
->-> roots=20
->-> are:<BR>x1=3D-0.250000 + 1.089725i , x2=3D-0.250000 - 1.089725i<BR>Pr ess =
->-> any key to=20
->-> continue*/</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>/*Please key in =
->-> the value of=20
->-> constant a,b and c for finding the roots of quadratic<BR>eq uation=20
->-> ax=B2+bx+c=3D0& nbsp; :4 2 5<BR>Your quadratic equation has two distinct =
->-> imaginary=20
->-> roots.&nbsp; Do you want to know<BR>the values of the imaginary roots=20
->-> (Y/N)?N<BR>q =3D N<BR>Thank you for using this computer<BR>Pre ss any key =
->-> to=20
->-> continue*/<BR>&nbsp;&nbsp ; <BR></STRONG></FONT></DIV></BODY></HTML>
->->
->-> ------=_NextPart_000_ 000E_01C5D424.9 4159240--
->->

[Red Dragon]

>>->-> any key to=20

->-> continue*/</STRONG></FONT></DIV>
->-> <DIV><FONT color=3D#000080 ><STRONG></STRONG></FONT>&nbsp;</DIV>
->-> <DIV><FONT face=3DArial color=3D#000080 size=3D2><STRON G>/*Please
key in =
->-> the value of=20
->-> constant a,b and c for finding the roots of quadratic<BR>eq uation=20
->-> ax=B2+bx+c=3D0& nbsp; :4 2 5<BR>Your quadratic equation has two
distinct =
->-> imaginary=20
->-> roots.&nbsp; Do you want to know<BR>the values of the imaginary
roots=20
->-> (Y/N)?N<BR>q =3D N<BR>Thank you for using this computer<BR>Pre ss any
key =
->-> to=20
->-> continue*/<BR>&nbsp;&nbsp ; <BR></STRONG></FONT></DIV></BODY></HTML>
->->
->-> ------=_NextPart_000_ 000E_01C5D424.9 4159240--
->->

Thank you Tim,
I absolutely have no idea of the problem until I saw your post to me. Not
even I dont like to read it, I am unable to read it.
I had purposely sent myself a returned copy of the mail and it was not like
this. The returned copy had not a single line of HTML code.
I suppose why this problem arises is because only readers with Outlook
Express get the mail in its perfect state. Others with different platform
will get it all in HTML.
Thanks for enlightening me.
Regards,
Khoon.

[Mark McIntyre]

On Tue, 18 Oct 2005 17:57:50 +0800, in comp.lang.c , "Red Dragon"
<ts*****@stream yx.com> wrote:

I dont quite understand the phrase "newline to terminate the line".

How many keys do you press, when you supply user input value of 1? One
key or two?
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt >

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
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[Red Dragon]

"Mark McIntyre" <ma**********@s pamcop.net> wrote in message
news:5t******** *************** *********@4ax.c om...

On Tue, 18 Oct 2005 17:57:50 +0800, in comp.lang.c , "Red Dragon"
<ts*****@stream yx.com> wrote:

I dont quite understand the phrase "newline to terminate the line".

How many keys do you press, when you supply user input value of 1? One
key or two?
--
Mark McIntyre

2 keys. One for 1 and One for Enter.
Can you kindly tell me is using 2 scanf() the best solution to the problem?
What is a better method?
Thanks
Rgds,
Khoon.