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Print table of a number entered by user

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Join Date: Jun 2009

Posts: 28

Reputation: seo2005 is an unknown quantity at this point

Solved Threads: 0

seo2005

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Light Poster

Print table of a number entered by user

27 Days Ago

Hi,

I would like to print the table of a number entered by the user like

2 *1=2
2*2=4
2*3=6

The code is

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
#include<stdio.h>
#include<conio.h>
 
main()
 
{
 
int i,num;
 
printf("Enter any number");
scanf("%d",&num);
 
for(i=1;i<=num;i++)
 
printf("%d*%d=%d",num,i,num*i);
 
getch();
 
}
#include<stdio.h>
#include<conio.h>

main()

{

int i,num;

printf("Enter any number");
scanf("%d",&num);

for(i=1;i<=num;i++)

printf("%d*%d=%d",num,i,num*i);

getch();

}

But this code is generating errors.

Last edited by peter_budo; 26 Days Ago at 5:09 am. Reason: Keep It Organized - For easy readability, always wrap programming code within posts in [code] (code blocks)

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Join Date: Sep 2009

Posts: 344

Reputation: dkalita will become famous soon enough

Solved Threads: 51

dkalita

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Posting Whiz

27 Days Ago

1> 28 posts and still no code tag

2> Try to practice int main(){} not just main() or void main()

3> Post the exact error that u are getting.
The code seems to be ok I think.

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Join Date: Oct 2009

Posts: 47

Reputation: Iam3R is an unknown quantity at this point

Solved Threads: 2

Iam3R

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Light Poster

26 Days Ago

The code is perfect and no errors execpt few things that shuold be avoided.
may be you want to print from

5 * 1 = 5
5 * 2 = 10
5 * 3 = 15
:
:
5 * 10 = 50

here is your code with slight modifications and proper posting:
follow this next time

 Help with Code Tags 
 C Syntax (Toggle Plain Text) 
#include<stdio.h>
 
int main()
{
        int i, num;
        printf( " Enter any number " );
        scanf( "%d", &num );
        for( i=1 ;i <=10; i++ )
            printf( "  \t %d * %d = %d \n ", num, i, num*i ) ;
        return 0;
}
#include<stdio.h>

int main()
{
        int i, num;
        printf( " Enter any number " );
        scanf( "%d", &num );
        for( i=1 ;i <=10; i++ )
            printf( "  \t %d * %d = %d \n ", num, i, num*i ) ;
        return 0;
}