#include<stdio.h>
void main()
{ int i=1,j=1,n=9;
printf("enter the char");
scanf("%d",&n);
while(i<=n)
{
while(j<=i)
{
printf("%d/0",i);
j++;
}
printf("/n");
i++;
}
}
OUTPUT
1
22
333
4444
55555
#include<stdio.h>
void main()
{ int i=1,j=1,n=9;
printf("enter the char");
scanf("%d",&n);
while(i<=n)
{
while(j<=i)
{
printf("%d/0",i);
j++;
}
printf("/n");
i++;
}
}
OUTPUT
1
22
333
4444
55555
Jump to Postwhat do u mean by 'enter the char' and read it as an integer.
Jump to Post/0,/n are wrong
use \0 and \n.
hope u r aware about it, but by a mistake??
what do u mean by 'enter the char' and read it as an integer.
/0,/n are wrong
use \0 and \n.
hope u r aware about it, but by a mistake??
thanks dude,but though the mistakes uve told arent syntax errors arent they
i ment "im not able to get oput"
in the sense i am not able to execute the program
and during compilation it gives no error
I think this code helps you
#include<stdio.h>
void main()
{ int i=1,j=1,n=9;
printf("enter the No");
scanf("%d",&n);
while(i<=n)
{
while(j<=i)
{
printf("%d\0",i);
j++;
}
printf("\n");
i++;
j=1;
}
}
Do u expect this?
@hsetaknev
line 4:
void main()
main always returns a int value so it must be
int main()
line 5:
when you are asking to the user for the value of n then
no need to initialize n=9.
line 13:
could'nt unserstand the need of /0 here even its \0.. no need for it.
line 16:
Its '\n' not '/n'.
Line 19:
place
return 0;
Post your code in proper format..
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