Simple output problem

[RadiationX]

I have to solve the following problem:Write a program that accepts two
integers, and determines if the second is a factor (is evenly divisible
into) the first. Here is the code i have so far.


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

int main()
{
int a; //dividend
int b; //divisor
int c;
c==0;

printf("<------The Factoring Machine------->\n");
printf("\nPleas e enter an integer followed by\n");
printf("a space and then the possible factor integer.\n");
printf("For exampe: [a b] is the correct format.\n");
scanf("%d%d", &a, &b);
c == a/b;


if(a%b==0)
{
printf("\n %d is a factor of %d\n", b, a);
printf("\nThe other factor is %d\n", c);

}

else
{
printf("\n%d is not a factor of %d\n",b,a);
}
system("PAUSE") ;
return 0;
}

why is the %d in the second printf not outputting the value of variable
c?
This should be the other factor.

[xyombie]

Re: Simple output problem

The problem is in the line that reads "c == a/b;". Using two equals is
a comparison (boolean condition). It's asking "is c equal to a/b?",
returning either a zero or one for true or false, but doesn't save this
value anywhere.

What you probably meant to do is "c = (int) a/b;" which will take the
value of a/b and place it into the variable c. Also, because a/b is
returning a double, it's probably a good idea to tell the compiler to
cast it back to an integer because c is an integer. Without the (int),
you might get a compiler warning depending on the compiler.

Good luck.


RadiationX wrote:[color=blue]
> I have to solve the following problem:Write a program that accepts two
> integers, and determines if the second is a factor (is evenly divisible
> into) the first. Here is the code i have so far.
>
>
> #include <stdio.h>
> #include <stdlib.h>
>
> int main()
> {
> int a; //dividend
> int b; //divisor
> int c;
> c==0;
>
> printf("<------The Factoring Machine------->\n");
> printf("\nPleas e enter an integer followed by\n");
> printf("a space and then the possible factor integer.\n");
> printf("For exampe: [a b] is the correct format.\n");
> scanf("%d%d", &a, &b);
> c == a/b;
>
>
> if(a%b==0)
> {
> printf("\n %d is a factor of %d\n", b, a);
> printf("\nThe other factor is %d\n", c);
>
> }
>
> else
> {
> printf("\n%d is not a factor of %d\n",b,a);
> }
> system("PAUSE") ;
> return 0;
> }
>
> why is the %d in the second printf not outputting the value of variable
> c?
> This should be the other factor.[/color]

[xyombie]

Re: Simple output problem

I also just reallized, you have the same problem in the line "c==0; ".
There is no need to initialize c, but if you wanted to, you would use
"c=0;".

As a reminder, double equal "==" is for comparison, while single "=" is
for setting a variable equal to another variable or equation. When you
mix this up, it can be a difficult bug to find because the compiler
isn't going to complain because it is valid C syntax.

[RadiationX]

Re: Simple output problem

that fixed it. thanks..

[abdur_rab7@yahoo.co.in]

Re: Simple output problem


RadiationX wrote:[color=blue]
> int a; //dividend
> int b; //divisor
> int c;
> c==0;
>[/color]

To avoid future errors, correct the initialization statement c==0; to
c=0;

---
Best Regards,
Abdur

[Vladimir S. Oka]

Re: Simple output problem

xyombie wrote:[color=blue]
> RadiationX wrote:[color=green]
>> I have to solve the following problem:Write a program that accepts
>> two integers, and determines if the second is a factor (is evenly
>> divisible into) the first. Here is the code i have so far.
>>
>>
>> #include <stdio.h>
>> #include <stdlib.h>
>>
>> int main()
>> {
>> int a; //dividend
>> int b; //divisor
>> int c;
>> c==0;
>>
>> printf("<------The Factoring Machine------->\n");
>> printf("\nPleas e enter an integer followed by\n");
>> printf("a space and then the possible factor integer.\n");
>> printf("For exampe: [a b] is the correct format.\n");
>> scanf("%d%d", &a, &b);
>> c == a/b;
>>
>>
>> if(a%b==0)
>> {
>> printf("\n %d is a factor of %d\n", b, a);
>> printf("\nThe other factor is %d\n", c);
>>
>> }
>>
>> else
>> {
>> printf("\n%d is not a factor of %d\n",b,a);
>> }
>> system("PAUSE") ;
>> return 0;
>> }
>>
>> why is the %d in the second printf not outputting the value of
>> variable c?
>> This should be the other factor.[/color][/color]

Please don't top post. I corrected it this time...
[color=blue]
>
> The problem is in the line that reads "c == a/b;". Using two equals
> is a comparison (boolean condition). It's asking "is c equal to
> a/b?", returning either a zero or one for true or false, but doesn't
> save this value anywhere.[/color]

You probably meant the correct thing, but your expression of it may be
confusing (see the order of true and false vis order of zero and one in
your sentence). A logical operation in C yields 1 if condition is true,
and 0 if it is false -- not the other way around, as your post implies.
[color=blue]
>
> What you probably meant to do is "c = (int) a/b;" which will take the
> value of a/b and place it into the variable c. Also, because a/b is
> returning a double, it's probably a good idea to tell the compiler to[/color]

`a/b` is _not_ returning a double. It returns an int, after
performing /integer/ division (any fractional part is discarded).
[color=blue]
> cast it back to an integer because c is an integer. Without the
> (int), you might get a compiler warning depending on the compiler.[/color]

There's no need for a cast here. All variables are of the type int, as
is the result of the division. What _is_ advisable though is to check
for division by zero (obviously before it is attempted), and do
something sensible if it's detected, e.g.:

if (0 == b)
{
fprintf(stderr, "Division by zero attempted!\n");
return EXIT_FAILURE;
}

Or ask user to try and input a legal value for b.

Cheers

Vladimir

--
What is mind? No matter.
What is matter? Never mind.
-- Thomas Hewitt Key, 1799-1875

[Mike Wahler]

Re: Simple output problem


"RadiationX " <heavyreader@gm ail.com> wrote in message
news:1138333690 .096638.186550@ g43g2000cwa.goo glegroups.com.. .[color=blue]
>I have to solve the following problem:Write a program that accepts two
> integers, and determines if the second is a factor (is evenly divisible
> into) the first. Here is the code i have so far.
>
>
> #include <stdio.h>
> #include <stdlib.h>
>
> int main()
> {
> int a; //dividend
> int b; //divisor
> int c;
> c==0;[/color]

c = 0;

or simply replace above two lines with:

int c = 0;

Look up the difference between = and == (they're
two distinct operators with different meanings.

[color=blue]
> printf("<------The Factoring Machine------->\n");
> printf("\nPleas e enter an integer followed by\n");
> printf("a space and then the possible factor integer.\n");
> printf("For exampe: [a b] is the correct format.\n");
> scanf("%d%d", &a, &b);[/color]

Be very careful with scanf(). See the FAQ for details.
[color=blue]
> c == a/b;[/color]

Again, wrong operator.

c = a / b;
[color=blue]
> if(a%b==0)
> {
> printf("\n %d is a factor of %d\n", b, a);
> printf("\nThe other factor is %d\n", c);
>
> }
>
> else
> {
> printf("\n%d is not a factor of %d\n",b,a);
> }
> system("PAUSE") ;
> return 0;
> }
>
> why is the %d in the second printf not outputting the value of variable
> c?[/color]

Actually it was. As your code is written, the only possible values
for 'c' were zero and one.
[color=blue]
> This should be the other factor.[/color]

-Mike

[Arndt Jonasson]

Re: Simple output problem


"xyombie" <brianblank@gma il.com> writes:[color=blue]
> I also just reallized, you have the same problem in the line "c==0; ".
> There is no need to initialize c, but if you wanted to, you would use
> "c=0;".
>
> As a reminder, double equal "==" is for comparison, while single "=" is
> for setting a variable equal to another variable or equation. When you
> mix this up, it can be a difficult bug to find because the compiler
> isn't going to complain because it is valid C syntax.[/color]

If you tell it to, it will:

% cat divv.c
int main()
{
int a,b,c;
a = b = c = 1;
c == a/b;
return 0;
}
% gcc -W divv.c
divv.c: In function `main':
divv.c:5: warning: statement with no effect

[Flash Gordon]

Re: Simple output problem

Arndt Jonasson wrote:[color=blue]
> "xyombie" <brianblank@gma il.com> writes:[color=green]
>> I also just reallized, you have the same problem in the line "c==0; ".
>> There is no need to initialize c, but if you wanted to, you would use
>> "c=0;".
>>
>> As a reminder, double equal "==" is for comparison, while single "=" is
>> for setting a variable equal to another variable or equation. When you
>> mix this up, it can be a difficult bug to find because the compiler
>> isn't going to complain because it is valid C syntax.[/color][/color]

Make that isn't required to. Some will.
[color=blue]
> If you tell it to, it will:[/color]

<snip>

No, it's not that it will, only that some do. There is no requirement to
produce a diagnostic since as xyombie says it is perfectly valid, and
I'm sure there are compilers that won't warn you.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.

[Nelu]

Re: Simple output problem

xyombie wrote:[color=blue]
> The problem is in the line that reads "c == a/b;". Using two equals is
> a comparison (boolean condition). It's asking "is c equal to a/b?",
> returning either a zero or one for true or false, but doesn't save this
> value anywhere.
>
> What you probably meant to do is "c = (int) a/b;" which will take the
> value of a/b and place it into the variable c. Also, because a/b is
> returning a double, it's probably a good idea to tell the compiler to
> cast it back to an integer because c is an integer. Without the (int),
> you might get a compiler warning depending on the compiler.[/color]
Please, don't top post.
--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Nelu]

Re: Simple output problem

xyombie wrote:[color=blue]
> I also just reallized, you have the same problem in the line "c==0; ".
> There is no need to initialize c, but if you wanted to, you would use
> "c=0;".
>
> As a reminder, double equal "==" is for comparison, while single "=" is
> for setting a variable equal to another variable or equation. When you
> mix this up, it can be a difficult bug to find because the compiler
> isn't going to complain because it is valid C syntax.
>[/color]
(Don't top post from the previous message)
Please quote some context, even if it's your post.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Nelu]

Re: Simple output problem

RadiationX wrote:[color=blue]
> that fixed it. thanks..
>[/color]
That fixed what? Thanks for what?
I'm glad it got fixed and that you are thanking someone.
Please let us know what you're talking about.
QUOTE SOME CONTEXT.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Arndt Jonasson]

Re: Simple output problem


Flash Gordon <spam@flash-gordon.me.uk> writes:[color=blue]
> Arndt Jonasson wrote:[color=green]
> > If you tell it to, it will:[/color]
>
> <snip>
>
> No, it's not that it will, only that some do. There is no requirement
> to produce a diagnostic since as xyombie says it is perfectly valid,
> and I'm sure there are compilers that won't warn you.[/color]

To be completely clear, I ought to have written, "if there is a way to
tell the compiler, then if you tell it to, it will." - I suppose I
made the impression of saying that every compiler works like 'gcc'.

[Dave Thompson]

Re: Simple output problem

On Fri, 27 Jan 2006 06:09:14 +0000 (UTC), "Vladimir S. Oka"
<novine@btopenw orld.com> wrote:
[color=blue]
> xyombie wrote:[/color]
<snip>[color=blue][color=green]
> > What you probably meant to do is "c = (int) a/b;" which will take the
> > value of a/b and place it into the variable c. Also, because a/b is
> > returning a double, it's probably a good idea to tell the compiler to[/color]
>
> `a/b` is _not_ returning a double. It returns an int, after
> performing /integer/ division (any fractional part is discarded).
>[/color]
Right. And even if a and b were floating-point, (int) a / b would
convert a to int (if in range, else Undefined Behavior), then convert
back to floating-point because of the usual arithmetic conversions for
computational operators, do the division in floating-point giving a
floating-point result, then implicitly convert to int (again if in
range and assuming c is int).

To explicitly force an integer result from flt-pt; c = (int) (a/b);

To explicitly do integer division: c = (int)a / (int)b;

(Or similarly for long, long long, unsigned, etc.)

- David.Thompson1 at worldnet.att.ne t