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#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?
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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 |
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 |
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