Thread: case, menu system, loops infinitely on invalid input

I am having a problem with a menu system, it's a sample menu system that I am putting together to make sure it will work, theoretical reasons, just trying to learn.

When an invalid value is entered it will go into an infinite loop and the program has to be forcefully terminated.

Can someone explain to me how to prevent this.

Code:

/* using an infinite loop and switch statement
   to implement a menu system */
#include <stdio.h>
#include <stdlib.h>

#define DELAY 150000

int menu(void);
void delay(void);

int main()
{

    while (1)
    {
        /* get user's selection and branch based on the input */

        switch(menu())
        {
            case 1:
            {
                puts("\nExecuting choice 1");
                delay();
                break;
            }
            case 2:
            {
                puts("\nExecuting choice 2");
                delay();
                break;
            }
            case 3:
            {
                puts("\nExecuting choice 3");
                delay();
                break;
            }
            case 4:
            {
                puts("\nExecuting choice 4");
                delay();
                break;
            }
            case 5:  /* exit program */
            {
                puts("\nExiting program now...\n");
                delay();
                exit(0);
            }
            default:
            {
                puts("\nInvalid choice, try again");
                delay();
            }
        }/* end of switch */

    } /* end of while */

}

/* displays a menu and inputs user selection */
int menu(void)
{
    int reply;

    puts("\nEnter 1 for task A");
    puts("Enter 2 for task B");
    puts("Enter 3 for task C");
    puts("Enter 4 for task D");
    puts("Enter 5 to exit program");

    scanf("%d", &reply);

    return reply;
}

void delay(void)
{
    long x;
    for (x = 0; x < DELAY; x++)
        ;
}

Thanks in advance.

I finally realized what was happening, and found information about converting strings to integers.

Ended up using this code for my menu() function and it does the trick. Thanks for pointing me in the right direction.

Code:

int menu(void)
{
    int reply;
    char buf[BUFSIZ];

    puts("\nEnter 1 for task A");
    puts("Enter 2 for task B");
    puts("Enter 3 for task C");
    puts("Enter 4 for task D");
    puts("Enter 5 to exit program");

    if(fgets(buf, sizeof(buf), stdin) != NULL)
    {
        reply = atoi(buf);
    }

    return reply;
}

You can do it that way or you can use fgetc() and make the numbers in your cases in to characters ... as in case '1':

Originally Posted by quzah

Code:

while( invalidinput == true );

Quzah.

LOL... a menu that only works on bad entries... I'm thinking ... while (invalidinput == false);

But yes, there is no reason --and little sense-- in splitting an obvious single result into two pieces. He should put the whole thing in main.

Originally Posted by ModeSix

Apparently I am not understand the implementation of fgetc()

The single most important skill a programmer can develop is ... looking "stuff" up...
The second most important skill is learning to read error messages... the error output you posted told you fgetc() expected a file pointer.

Originally Posted by quzah

No, I had this in mind:

No silly... like this...

Code:

invalidinput = false;
do
{
    printf( "enter 1, everything else is invalid: " );
    fflush( stdout )

    switch( fgetc( stdin ) )
    {
        case '1':
            printf( "yay\n" );
        break;
        default:
            invalidinput = true;
            printf( "invalid input. try entering 1 again!\n" );
        break;
    }
} while( invalidinput == false );