While Loop skipping scanf for it's condition.

Question

I can't see why does the While loop just speed away and skipping the scanf for the char? It would not even ask for my input and just loop like there's no tomorrow.

#include <stdio.h>


int main()
{
    int number;
    int multiply, ans;
    char choice;

    printf("-------------------------------------");
    printf("\n      MULTIPLICATION TABLE           ");
    printf("\n-------------------------------------");


    do
    {

         printf("\nEnter an integer number:");
         scanf("%d", &number);


        printf("\nMultiplication of %d is :-\n", number);
        printf("\n");

        for(multiply=1; multiply<11; multiply++){
            ans = number * multiply;
            printf(" %d", ans);
        }

        printf("\n");
        printf("\nWould you like to continue? [Y] for Yes,[N] for no : ");
        scanf("%c", &choice);
        printf("\n");

    } 
    while(choice='Y');

    printf("Thank You");
    return 0;

}

score 2 · Accepted Answer · 2012-12-24 11:26:51Z

2

scanf() doesn't do what you think it does (newline characters, buffering, etc.). It's preferred to use fgetc():

choice = fgetc(stdin);

For the same reason, you need to get rid of the trailing newline that

scanf("%d", &number");

leaves in the standard input buffer. To fix this, insert

fgetc(stdin);

after that particular call to scanf().

Also, C is not Pascal. The equality comparison operator - and the condition - you're looking for is

while (choice == 'Y')

the single equation mark denotes an assignment.

edited Dec 24, 2012 at 11:26

answered Dec 23, 2012 at 7:23

user529758

Sign up to request clarification or add additional context in comments.

4 Comments

user529758 Over a year ago

@DCoder And fixing that doesn't fix the erroneous assumptions about scanf().

Amin Husni Over a year ago

Changed it but still skipping the input part.

user529758 Over a year ago

@user1924648 what kind of insane input data are you using? Tried inserting another call to fgetc() after scanf("%d", &number);?

user529758 Over a year ago

@user1924648 Yep, it does the trick, I just tried it as well.

Maxwell Evans · Accepted Answer · 2012-12-23 07:26:37Z

2

It has been a long time since I programmed in that language, but at a glance, you have:

while(choice='Y');

instead of:

while(choice=='Y');

== compares, = sets equal to. So the while loop is actually not checking the condition you're trying to set.

answered Dec 23, 2012 at 7:26

Maxwell Evans

1372 silver badges10 bronze badges

1 Comment

user529758 Over a year ago

"I've fix that part but it still won't wait for me to put an input. – user1924648 1 min ago"

Yogendra Singh · Accepted Answer · 2012-12-23 07:23:08Z

1

I think you need to use == operator for the comparison in while condition check as:

   while(choice=='Y');

Currently you are using = operator, which is assigning Y to choice variable.

answered Dec 23, 2012 at 7:23

Yogendra Singh

34.5k7 gold badges66 silver badges73 bronze badges

3 Comments

Amin Husni Over a year ago

I've fix that part but it still won't wait for me to put an input.

user529758 Over a year ago

@user1924648 Use fgetc(), really.

Amin Husni Over a year ago

Used fgetc() and now it's going out of the loop without waiting for me to enter anything. It's working now, I just inserted an additional fgetc after getting the int.

Collectives™ on Stack Overflow

While Loop skipping scanf for it's condition.

3 Answers 3

4 Comments

1 Comment

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

4 Comments

1 Comment

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related