Thread: Making an Algorithm repeat itself

You need to flush the input buffer before you prompt for the repeat or not. Or as I like to say, turn off the faucet.


Quzah.

I don't really understand what that is supposed to accomplish. I'm trying to get the entire program to repeat if the user selects 'y'. Here's what I've been trying to do.

(not actual code)

Code:

while (choice == 'y')
{
funct(1) Calculation
funct(2) Answer
funct(3) print answer

printf("Would you like to find another answer (y or n)? ");
scanf("%c", &choice);
}
printf("Goodbye.");
return 0;

I'm not sure what this flushing of the input buffer does or actually how to do it. I was actually hoping for a simpler way to do it.

No... here is an idea. Maybe it's best that you click on the link provided before saying you don't get it.

You flush the input buffer because when you try to scan your character for the repeat statement at the end, the program will take what's already in the input buffer (usually a newline character). You need to clear the buffer before you prompt for that input otherwise it will appear to have skipped over it.

Ok, then use a scanf() or getch() to scan in the junk character or something.

Of course, it will help if choice is set to 'y' before you reach the loop. Another way to do this will be to use the do {} while () construct, as in

Code:

do
{
    funct(1) Calculation
    funct(2) Answer
    funct(3) print answer

    printf("Would you like to find another answer (y or n)? ");
    scanf("%c", choice);
} while (choice == 'y');
printf("Goodbye.");
return 0;

Of course, this doesn't fix the fact that scanf() doesn't consume all of the input from the console, but it's actually what you were asking rather than the answer to a question that you'll have once you solve the problem you're having now.

Originally Posted by mvd1212

scanf("%c", %choice);

Originally Posted by filker0

scanf("%c", choice);

Boy you guys are having a heckuva time with that one today...

Code:

char choice;
...
scanf("%c", &choice );

[edit]
LMAO. I forgot the 'choice' in my smart ass reply and had to edit it in. But hey, I got the colored &. But in my defense, I have numerous alcoholic beverages as witnesses.
[/edit]

Quzah.

Originally Posted by quzah

Boy you guys are having a heckuva time with that one today...

Code:

char choice;
...
scanf("%c", &choice );

Oops... I cut/pasted his code sample, then when I saw the %, I went in and edited it out. Yes, you need the & for a %c. I had considered suggesting that he do a %s, make choice an array, and use

Code:

while (choice[0] == 'y');

Typing as someone who's been coding in C since 1981 and has written more than one printf() implementation, I hang my head in shame at my omission.

Hmmm... I suppose you could read "choice" in a loop until you got either "y" or "n", as in

Code:

choice = '\0';
while (choice != 'y' && choice != 'n')
{
    scanf("%c", &choice);
}

Eventually, it will find a 'y' or a 'n'. But that, of course, isn't really what you want. You could change your format string to "%c\n", as in

Code:

scanf("%c\n", &choice);

which will discard a trailing newline.

You may want to check your scanf() calls within the functions... If you're not consuming the newlines at the end of the input there, it explains why you still have them in the input stream when you get to the end of your loop.

Originally Posted by mvd1212

filker0... I love you.

Ahh... That's awkward... I'm already married. Glad to be of service, though.

Originally Posted by mvd1212

Problem solved.

The addition of \n to "scanf("\n%c", &choice)" completely fixed the problem.


filker0... I love you.

Course just adding a getch() or scanf() to junk the newline like I've been saying all the time would have worked too, but whatever. I'm used to being stepped on.