Thread: Repetition in do{}while

No. Your solution only works if they only enter "key""enter". If they enter more than one "key" before hitting enter, it'll loop for as many "key"s you press. Replace your getchar line with the while solution Thantos mentioned.

Remember, you have to code down to the level of the stupidest user.

Quzah.

i tried that after you implement that the program won't let you enter anything on the second time around i was just saying that the getchar() there is a quick solution

Originally Posted by prog-bman

i tried that after you implement that the program won't let you enter anything on the second time around i was just saying that the getchar() there is a quick solution

I can't understand what it is you're trying to say in that sentence. See how benificial it is when you use capitalization and punctuation?

At any rate, what's not "quick" about this solution? It takes 30 seconds to type the whole thing, and it works exactly as it should. (Considering EOF usually kills the entire program anyway, there's no need to bother testing for it, especially since the input is buffered anyway.)

Code:

#include <stdio.h>

int main( void )
{
        int c;
        do
        {
                printf("This is a menu.\n");
                printf("This is a prompt. > ");
                c = getchar( );
                while( getchar( ) != '\n' );

        } while ( c != 'Q' && c != 'q' );

        return 0;
}

Presto. Perfect menuing.

Quzah.

Not perfect menuing after the loop runs once you have to input your answer twice in order for it to get picked up by the switch

Good call go with fgets totally forgot about that :P

>I'm going to go read up on fgets


>and how to flush the input buffer
Yes, do. But before you do, commit to memory the fact that if you need to flush the input buffer, you've already done something wrong. 1) If you need to then your program is ill-designed and you should fix it rather than apply a Band-Aid and 2) The techniques for "flushing the input stream" still aren't a solid solution, if you consider them a solution at all.

Originally Posted by prog-bman

Not perfect menuing after the loop runs once you have to input your answer twice in order for it to get picked up by the switch

No it doesn't. Did you even run it? You don't have to enter anything twice.

There is no problem with my example. Run it and see. I dare you to break this using anything other than EOF to kill it (or some signal or what not). Just using standard input, you cannot break mine, and it does not give invalid entries, and does not make you do it twice.

The only thing it does "wrong" is that if you enter a valid choice as your first letter, followed by extra crap, it will take the first and kill the rest.

Code:

#include <stdio.h>
int main( void )
{
    int c;
    do
    {
        printf("Menu:\n");
        printf("1) foo \n");
        printf("2) bar \n");
        printf("Q) quit \n");
        printf("> ");

        c = getchar( );
        while( getchar() != '\n' );
        if( c == '1' )
            printf("you chose one\n");
        else if ( c == '2' )
            printf("you chose two\n");
        else if ( c == 'q' || c == 'Q' )
            printf("goodbye\n");
        else
            printf("try again");
    } while( c != 'q' && c != 'Q' );
    return 0;
}

The only "problem" is that it doesn't care if you enter a valid followed by invalid data, because it only cares about the first key. But worse case scenario, you have to make your choice because you entered the first key as invalid, which from the description, is what you want the menu to do anyway.

This does not make you enter anything twice. I don't know what crack you're smoking, but put the pipe down long enough to compile it and acknowledge the fact that you're wrong.

Quzah.

Quzah i wasn't talking about your program I was talking about his. Your method yours works pefectly fine for your example