issue with switch statement

HI there,

I implemented a menu with a switch statement with which one can navigate through the different options I offered, which coincide, in turn, with the cases on the switch statement.

I intended the line that contains 'default' for the menu to return to the initial state, that is where it asks me for the options I want to enter.

However instead of returning to the default state in which I can select the option that I want, the program terminates. Everything else works fine in the program... I thought about implementing a goto statement but this would be in violation of 'good programming practices' .... what can I do?

Also in order for the program to continue until the end of the process I removed the break; statements within the switch statement. Is this orthodox? I am thinking that perhaps I should have used if statements and do whiles... any suggestions?

Thank you

Code:

int main(int argc, char** argv) {

    int option;

    menu();


  printf("Enter option: "); // this is the state in which I want the program when finishing <<<HERE
      scanf("%d",&option);

  switch (option) {

      case 1:printf("Create employee\n");
    fill_staff(); // takes index from user 
    employee_create(); // fills an array with index above

      case 2: printf("Enter employee data\n");
    for (d = 1; d <= emp_num; d++)
    employee_data(); //enter data for the employees created

    case 3:printf("Print employee list\n");
    for (d = 1; d <= emp_num; d++)
    print_employee(); //prints the array of employees
   
    default: printf("\n\n\nEnter option: "); //Return back to the original state and continue choosing options -- THIS IS WHERE THE PROBLEM IS <<<
      scanf("%d",&option);
  }
 
    return (EXIT_SUCCESS);
}

your program ends after the switch block because there is no loop, a condition i in the switch is matched, or default is used, then the program flow will just continue beyond the end of the closing switch braces, ie exit.

you need to wrap it in a loop, use a while loop starting above the switch and ending after the close brace.
set the condition in the while loop to "while option is not equal to 'whatever'"
this way it will continue requesting options until number 5 or whatever you decide to exit on is pressed.
you need to cover possibility of invalid input in your scanf statements, which can also lead to infinite loops..

It's normally

Code:

case 1:
   // some stuff
   break;

Another way to look at doing the program:

Code:

do
{

printf("If you would like to do this, enter 1);
printf("if you would like to do that, enter 2);
printf("to exit the program, enter 0);
scanf("%i", &x);


if(x = 1)
....
if(x = 2)
....

}
while(x != 0);

I have no idea how "menu" works, so don't shoot me if you have to use a switch statement. :P

Thank you guys...!! I will try the suggestion with the loop statement right away..

BR,

blue

I re-wrote the code to include the suggestion...However what I get now is that when I select an option it is permanently stuck in that option. If I select 'create employee' which corresponds to case 1 then it permanently asks me to enter employees. Did I do the while statement right? How do I move on to other cases? I post the updated code below

Code:

int main(int argc, char** argv) {

    int option;

    menu();


  printf("Enter option: ");
      scanf("%d",&option);


  while (option!=0){

  switch (option) {

      case 1:printf("Create employee\n");
    fill_staff(); // takes index from user
    employee_create(); break;// fills an array with index above

      case 2: printf("Enter employee data\n");
      for (d = 1; d <= emp_num; d++)
    employee_data(); break;//enter data for the employees created

    case 3:printf("Print employee list\n");
    for (d = 1; d <= emp_num; d++)
    print_employee();break; //prints the array of employees

     case 4:printf("Exit program: ");
            option = 0;break;

    default: printf("\n\n\nEnter option: "); //Return back to the original state and continue choosing options
      scanf("%d",&option);break;
        }

  }
 
    return (EXIT_SUCCESS);
}

default: printf("\n\n\nEnter option: "); //Return back to the original state and continue choosing options
scanf("%d",&option);break;

Remove this and get scanf out of switch, in the end of while block. Also, case 4 is not needed and will not actually work since you would assign option a value of 0 and then immediately overwrite it with a user's response.

You need to get the option every time and based on that you are going to do the stuffs na.
So

Code:

while(option != 0)
{
         printf("Enter the option");
         scanf("%d",&option);
         switch(option)
         {
                  case 1:
                            //stuffs.....
                  case 2:
                           //stuffs...
                     .
                     .
                     .
                  case n:
                           //stuffs
                           break
                  default:
                          printf("Invalid option");
                          break;
           }
}

Another minor thing I see is that you just say:

Code:

int option;

You want to set it equal to something other than zero. It may work 99% of the time, but that 1% of the time may be the 1% that your professor grades your assignment.

Hi GL SAM,

I think I am understanding you wrong... By removing the default line out of the switch I get an error. '... main.c:118: error: 'default' label not within a switch statement...'

If this is what you meant? I paste the code which does not work and thereafter the one that does work...

Code:

int main(int argc, char** argv) {

    int option;

    menu();


    printf("Enter option: ");
    scanf("%d", &option);


    while (option != 0) {

        switch (option) {

            case 1:printf("Create employee\n");
                fill_staff(); // takes index from user
                employee_create(); break;// fills an array with index above

            case 2: printf("Enter employee data\n");
                for (d = 1; d <= emp_num; d++)
                    employee_data();break; //enter data for the employees created

            case 3:printf("Print employee list\n");
                for (d = 1; d <= emp_num; d++)
                    print_employee(); break;//prints the array of employees

            case 4:printf("Exit program: ");
                option = 0;break;


        }
        default: printf("\n\n\nEnter option: "); //Return back to the original state and continue choosing options <<< THIS IS MOVED OUT OF THE SWITCH STATEMENT
        scanf("%d", &option);break;
    }

    return (EXIT_SUCCESS);
}

If I, on the other side, remove the break; statements I get the desired result.... why is that? I paste the code that works below:

Code:

int main(int argc, char** argv) {

    int option;

    menu();


    printf("Enter option: ");
    scanf("%d", &option);


    while (option != 0) {

        switch (option) {

            case 1:printf("Create employee\n");
                fill_staff(); // takes index from user
                employee_create(); // fills an array with index above

            case 2: printf("Enter employee data\n");
                for (d = 1; d <= emp_num; d++)
                    employee_data(); //enter data for the employees created

            case 3:printf("Print employee list\n");
                for (d = 1; d <= emp_num; d++)
                    print_employee(); //prints the array of employees

            case 4:printf("Exit program: ");
                option = 0;

            default: printf("\n\n\nEnter option: "); //Return back to the original state and continue choosing options
                scanf("%d", &option);
        }

    }

    return (EXIT_SUCCESS);
}

Originally Posted by madmax2006

Another minor thing I see is that you just say:

Code:

int option;

You want to set it equal to something other than zero. It may work 99% of the time, but that 1% of the time may be the 1% that your professor grades your assignment.

I take notice ;-)

Just a bit of tweaking:

Code:

int main(int argc, char** argv) {

    int option = 1;

    menu();


    while (option != 0) {

        printf("Enter option: ");
        scanf("%d", &option);
        getchar();  //pulls off the remaining newline

        switch (option) {

            case 1:printf("Create employee\n");
                fill_staff(); // takes index from user
                employee_create(); break;// fills an array with index above

            case 2: printf("Enter employee data\n");
                for (d = 1; d <= emp_num; d++)
                    employee_data();break; //enter data for the employees created

            case 3:printf("Print employee list\n");
                for (d = 1; d <= emp_num; d++)
                    print_employee(); break;//prints the array of employees

            case 4:printf("Exit program: ");
                option = 0;break;


             default: printf("\n I don't have that option available, please try again");
        }
    }

    return (EXIT_SUCCESS);
}

You might want to consider something like this:

Code:

int option = menu();
while (option != OPTION_QUIT) {
    switch (option) {
        /* handle valid options */
    default:
        /* print error message concerning invalid option */
    }
    option = menu();
}

My idea is that handling the menu consists of two parts: printing it, and reading the menu option. Thus, menu() does both (possibly by calling other functions that handle the two individual parts), then returns the option read.

Now, instead of printing the menu in the default case, you print an error message. It is always the case that as long as the option is not an option to quit (i.e., the option matches the OPTION_QUIT constant), the menu will be printed and the next option read.

Originally Posted by laserlight

You might want to consider something like this:

Code:

int option = menu();
while (option != OPTION_QUIT) {
    switch (option) {
        /* handle valid options */
    default:
        /* print error message concerning invalid option */
    }
    option = menu();
}

My idea is that handling the menu consists of two parts: printing it, and reading the menu option. Thus, menu() does both (possibly by calling other functions that handle the two individual parts), then returns the option read.

Now, instead of printing the menu in the default case, you print an error message. It is always the case that as long as the option is not an option to quit (i.e., the option matches the OPTION_QUIT constant), the menu will be printed and the next option read.

Thank you much laser light as a matter of fact thank you all!

This seems a more sophisticated way to make the menu...

I guess that for this I would have to have two functions one to read and one to print... and perhaps even some parameters right?

Blue

i think your original menu() just displayed the options right? laserlight has shown that you could use the function to return the option chosen, thus this could be your 'read' function. Then you could create a 'write' function that carries option as an integer argument. then all the print functions you previously wrote in the switch cases could be carried out in that new function.

you could use the switch cases to just set a variable to show that the option is valid, then call the write function at the end based on that (use an if statement to control as you wont want to call write() if the default case is matched and option was invalid. un less you want write() to display your error messages too)

or just call write() in each switch case.