where the mistake is

[yuanshuaisd@yahoo.cn]

#include<stdio. h>
int main()
{
int n=1;
int average;
int num;
int sum;
for(n=1;num != 9999;n++){
printf("Enter a integer:\n");
scanf("%d",&num );

sum = sum + num;
}

average = sum / n;
printf("average is %d",average);

return 0;
}
why the program end in a wrong result

[osmium]

Re: where the mistake is

<yuanshuaisd@ya hoo.cnwrote:

#include<stdio. h>
int main()
{
int n=1;
int average;
int num;
int sum;
for(n=1;num != 9999;n++){
printf("Enter a integer:\n");
scanf("%d",&num );
>
sum = sum + num;
}
>
average = sum / n;

average has been declared as an integer. In general, there are decimal
results in averages so change average to a variable of type double. There
may be other problems as well, I didn't look for them.

printf("average is %d",average);
>
return 0;
}
why the program end in a wrong result

[AnonMail2005@gmail.com]

Re: where the mistake is

On Nov 3, 9:05 am, yuanshua...@yah oo.cn wrote:

#include<stdio. h>
int main()
{
        int n=1;
        int average;
        int num;
        int sum;
        for(n=1;num != 9999;n++){
        printf("Enter a integer:\n");
                scanf("%d",&num );
>
                sum = sum + num;
        }
>
    average = sum / n;
        printf("average is %d",average);
>
        return 0;}
>
why the program end in a wrong result

For one, the variables num and sum are not initialized.

HTH

[Juan Antonio Zaratiegui Vallecillo]

Re: where the mistake is

yuanshuaisd@yah oo.cn wrote:

#include<stdio. h>
int main()
{
int n=1;

'num' initialised to 1, and reinitialized later in the loop, you may
just intialise it once

int average;
int num;
int sum;

'sum' is not initialised

for(n=1;num != 9999;n++){
printf("Enter a integer:\n");
scanf("%d",&num );

'scanf' return value is not checked for errors

>
sum = sum + num;
}
>
average = sum / n;

'average' is integer, so the result will be such that:
result<= real average< result+1

printf("average is %d",average);
>
return 0;
}
why the program end in a wrong result

When you add anything to an undefined value (not initialised) the result
is undefined behaviour. That is, you may even get the result you
expected. Now. But not later, specially if you are showing your program
to another party.

Best regards,

Zara

[Christopher Dearlove]

Re: where the mistake is

<yuanshuaisd@ya hoo.cnwrote in message
news:0fdbc6bb-1c2f-411b-8eaf-90c322e931bf@v1 3g2000pro.googl egroups.com...

for(n=1;num != 9999;n++){

That's almost certainly not the loop you want. Apart from the confusion
of having num in the test, even if that is the test you want, note that you
add numbers until you enter a 9999 - but do add in that 9999.

[Juan Antonio Zaratiegui Vallecillo]

Re: where the mistake is

Juan Antonio Zaratiegui Vallecillo wrote:

yuanshuaisd@yah oo.cn wrote:

>#include<stdio .h>
>int main()
>{
> int n=1;

>
'num' initialised to 1, and reinitialized later in the loop, you may
just intialise it once

No, this comment is wrong.

>

> int average;
> int num;
> int sum;

>
'sum' is not initialised
>

> for(n=1;num != 9999;n++){

Shouldn't it be:
while ( num != 9999 ) {
?

> printf("Enter a integer:\n");
> scanf("%d",&num );

>
'scanf' return value is not checked for errors
>

>>
> sum = sum + num;
> }
>>
> average = sum / n;

>
'average' is integer, so the result will be such that:
result<= real average< result+1
>

> printf("average is %d",average);
>>
> return 0;
>}
>why the program end in a wrong result

>
When you add anything to an undefined value (not initialised) the result
is undefined behaviour. That is, you may even get the result you
expected. Now. But not later, specially if you are showing your program
to another party.
>
Best regards,
>
Zara

[Juha Nieminen]

Re: where the mistake is

yuanshuaisd@yah oo.cn wrote:

#include<stdio. h>
int main()
{
int n=1;
int average;
int num;
int sum;
for(n=1;num != 9999;n++){
printf("Enter a integer:\n");
scanf("%d",&num );
>
sum = sum + num;
}
>
average = sum / n;
printf("average is %d",average);
>
return 0;
}
why the program end in a wrong result

Because to terminate the program you are entering 9999, which is added
to the sum?

[Fred]

Re: where the mistake is

On Nov 3, 11:55 am, Juha Nieminen <nos...@thanks. invalidwrote:

yuanshua...@yah oo.cn wrote:

#include<stdio. h>
int main()
{
   int n=1;
   int average;
   int num;
   int sum;
   for(n=1;num != 9999;n++){
   printf("Enter a integer:\n");
           scanf("%d",&num );

>

           sum = sum + num;
   }

>

    average = sum / n;
   printf("average is %d",average);

>

   return 0;
}
why the program end in a wrong result

>
  Because to terminate the program you are entering 9999, which is added
to the sum?- Hide quoted text -
>
- Show quoted text -

Also, n is incorrect when the loop terminates.
--
Fred

[asterisc]

Re: where the mistake is

On Nov 3, 4:05 pm, yuanshua...@yah oo.cn wrote:

#include<stdio. h>
int main()
{
        int n=1;
        int average;
        int num;
        int sum;
        for(n=1;num != 9999;n++){

Reading 'num' without being initialized yield undefined behavior.