Thread: Quick question

Why is my switch statement looping through all scenarios and printing each case.The program works correctly, but it should break after first case if I enter a number.??????.......This is just the part of the code I'm having a problem with.............Thanks.

Code:

/* while user does not enter E */
   while ( c != 'E') { 
      switch( c ) { 

         /* enqueue value */
         case 'A':
            printf( "Enter a number: " );
            scanf( "\n%d", &item );
            enqueue( &headPtr, &tailPtr, item );
            printQueue( headPtr );
            break;

         /* dequeue value */
         case 'R':

            /* if queue is not empty */
            if ( !isEmpty( headPtr ) ) { 
               item = dequeue( &headPtr, &tailPtr );
               printf( "%d has been dequeued.\n", item );
            } /* end if */

            printQueue( headPtr );
            break;

           default:
            printf( "Invalid choice.\n\n" );
            instructions();
            break; 

      } /* end switch */

      printf( "? " );
      scanf( "%c", &c );
   } /* end while */

   printf( "End of run.\n" );
   
   return 0; /*

It does the case first, but also displays the default senario also, then loops again....Why is it displaying the default part......Maybe something wrong with while statement??????.Thanks for any help.

I think you have forgotten that \n is a perfectly good character, and therefore every other time through the loop c is enter-key rather than your choice.

At first glance, you aren't updating C after each switch statement. For example, you enter 'A', and you add a number to the list, then the program breaks and loops back around to the A case again. You need to ask the user to enter C again.

Good advice.Thanks.....I'll give it another look.........thanks...

I don't know....Its just not breaking at the break statement as it should.

In his post, tabstop has told you everything you need to know to fix the source code.