#include<stdio.h> #include<conio.h> void main() {clrscr(); char

#include<stdio.h>
#include<conio.h>
void main()
{clrscr();
char another='y';
int num;
for(;another=='y';)
{
printf("Enter a number");
scanf("%d",&num);
printf("squre of %d is %d",num,num*num);
printf("\nwant to enter another number y/n");
scanf("%c",&another);
}
getch();
}
the above code runs only one time.on entering 'y' the
screen disappeares.what can i do?

Question Posted / amar shukla

3 Answers
6306 Views
I also Faced
E-Mail Answers

Answers were Sorted based on User's Feedback

#include<stdio.h> #include<conio.h> void main() {clrscr(); char another='y..

Answer / shahid sayyad

#include<stdio.h>
#include<conio.h>
void main()
{clrscr();
char another='y';
int num;
for(;another=='y';)
{
printf("Enter a number");
scanf("%d",&num);
printf("squre of %d is %d",num,num*num);
printf("\nwant to enter another number y/n");
fflush();
scanf("%c",&another);
}
getch();
}
if you enter any key it act as char and store in
"another",so before scanning we have to flush all the keys
hence fflush is used.

Is This Answer Correct ?

4 Yes

0 No

#include<stdio.h> #include<conio.h> void main() {clrscr(); char another='y..

Answer / sight

You just remove the & in the last scanf statement
But also this loop will continue going on(infinite) b'coz
here no ending point.....
BTW if u remove those & it will definatly wait for entering
a charecter....

Is This Answer Correct ?

4 Yes

1 No

#include<stdio.h> #include<conio.h> void main() {clrscr(); char another='y..

Answer / jyoti goyal

Just replace %c by %s in the last scanf.

Is This Answer Correct ?

0 Yes

1 No

Post New Answer