Wont return to menu() function

So I've got another problem with a program. It's not really that big of a deal, but I can't seem to figure out why it's not working. I searched, but I couldn't find anything related to this. I've got a geometry calculation program, you decide which shape you want to calculate for, and then the actual calculation. Then you input the values necesary for the calculation. The program calculates, and outputs the answer. Finally it returns to the menu, incase you need to do more calculations. Everything works but returning to the menu() function, and I can't figure out why.
Any help is greatly appreciated. Thanks!

Code:

---

#include <iostream>
#include <string>
#include <math.h>

//Funtion prototypes.
void menu();
int which_calc();
void calc_circle();
void calc_rectangle();
void calc_triangle();

using namespace std;

int main()
{
    menu();
    system("pause");
    return (0);
}

void menu()
{
    int choice = 0;
    do
    {
        cout << "Which formula would you like to use?\n\n";
        cout << "  1.  Calculate for a circle\n";
        cout << "  2.  Calculate for a rectangle\n";
        cout << "  3.  Calculate for a triangle\n";
        cout << "  4.  Quit\n";
        cin >> choice;
        if (choice < 1 || choice > 4)
          cout << "Please enter a correct menu number\n";
    }while (choice < 1 || choice > 4);
   
    switch(choice)
    {
        case 1: cout << "You chose calculate for a circle\n\n";
                calc_circle();
                break;
        case 2: cout << "You chose calculate for a rectangle\n\n";
                calc_rectangle();
                break;
        case 3: cout << "You chose calculate for a triangle\n\n";
                calc_triangle();
                break;
        case 4:
                break;   
    }//!End of switch/case
    return;
}

/*
Function whichCalc() gets rid of several lines of repeating code
by asking which calculation they want and returning int choice.
*/
int which_calc()
{
    int choice = 0;
    do
    {
        cout << "Which calculation would you like?\n\n";
        cout << "  1. Circumference\n";
        cout << "  2. Area\n";
        cin >> choice;
        if (choice < 1 || choice > 2)
        {
          cout << "Please enter a correct menu number\n";
        }
    }while(choice < 1 || choice > 2);
    return(choice);
}

void calc_circle()
{
    int choice = which_calc();
    double radius, area, circumference;
    const double PI = 3.141592654;
   
    cout << "Please enter radius: ";
    cin >> radius;
   
    if (choice == 1)
    {
        cout << "The circumference of a circle with a radius of " << radius
              << " is " << (2 * PI * radius) << endl;
    }
    else
    {
        cout << "The area of a circle with a radius of " << radius
              << " is " << (PI * PI * radius) << endl;
    }
    menu();  //This is where I want it to return to the original menu.
}

---

Maybe callee can't call its caller. Make a loop in menu(). Run your program in debug mode.

[EDIT]
In your code functions actually don't return to menu(), they call it.

What's the compiler you are using? Your code works as you expected with MinGW 3.4.5.

The only change I had to make was to comment the calls to calc_triangle() and calc_rectangle() since these functions weren't defined. But tis has no effect on your problem.

The OP wants to call menu infinitely, I think.

Try something like this:

Code:

---

// in main()
while(!menu());

---

and make menu() return 0 to continue looping, 1 to quit.

Quote:

---

Originally Posted by System_159

Code:

---

    menu();  //This is where I want it to return to the original menu.

---

---

It actually calls menu() and as such will only leave when the option chosen is 4. So i'm a little baffled.

Well what if you wanted to calculate multiple triangles, then where would you be?

I think the OP's looking for a sort of loop around the existing code, which could be implemented as a loop in menu() or as a loop in main() based on the return value of menu().

Well, true.

I've been assuming he has the other functions defined. As is, the code will give a link-time error.

Read the original post.

Quote:

---

I've got a geometry calculation program, you decide which shape you want to calculate for, and then the actual calculation. Then you input the values necesary for the calculation. The program calculates, and outputs the answer. Finally it returns to the menu, incase you need to do more calculations. Everything works but returning to the menu() function, and I can't figure out why.

---

It definitely sounds like it compiled. It sounds a lot like they just need to call menu() again.

If you continue to run that code it will result in a stack overflow since the functions being called never return and menu never actually returns, thus the old stack frames are still in effect.

dwks is right. I just want to have the menu reappear until the user inputs "4" to quit.
I tried using return(menu()); at the end of the circle function. I also changed the actual menu function to return the corresponding function instead of just calling it. Then I tried clearing the input at the beginning of menu(). None of that worked though. :(

Quote:

---

Try something like this:

Code:

---

// in main()
while(!menu());

---

and make menu() return 0 to continue looping, 1 to quit.

---

I'm not sure what toyu mean by this dwks

Thanks for all the ideas so far, though.

Code:

---

void menu()
{
    for(;;){
        int choice = 0;
        do
        {
            cout << "Which formula would you like to use?\n\n";
            cout << "  1.  Calculate for a circle\n";
            cout << "  2.  Calculate for a rectangle\n";
            cout << "  3.  Calculate for a triangle\n";
            cout << "  4.  Quit\n";
            cin >> choice;
            if (choice < 1 || choice > 4)
              cout << "Please enter a correct menu number\n";
        }while (choice < 1 || choice > 4);
   
        switch(choice)
        {
                case 1: cout << "You chose calculate for a circle\n\n";
                      calc_circle();
                      break;
                case 2: cout << "You chose calculate for a rectangle\n\n";
                          calc_rectangle();
                      break;
                case 3: cout << "You chose calculate for a triangle\n\n";
                        calc_triangle();
                      break;
                case 4:
                      return;   
              }//!End of switch/case
    }
    return;
}

---

And erase menu() at the end of calculation functions.

Quote:

---

And erase menu() at the end of calculation functions.

---

This is extremely important.

My fix was the following:

Code:

---

#include <iostream>
#include <string>
#include <math.h>

//Funtion prototypes.
int menu();
int which_calc();
void calc_circle();
void calc_rectangle();
void calc_triangle();

using namespace std;

int main()
{
    // while(menu()) {} // edited
    while(!menu()) {}
    system("pause");
    return (0);
}

int menu()
{
    int choice = 0;
    do
    {
        cout << "Which formula would you like to use?\n\n";
        cout << "  1.  Calculate for a circle\n";
        cout << "  2.  Calculate for a rectangle\n";
        cout << "  3.  Calculate for a triangle\n";
        cout << "  4.  Quit\n";
        cin >> choice;
        if (choice < 1 || choice > 4)
          cout << "Please enter a correct menu number\n";
    }while (choice < 1 || choice > 4);
   
    switch(choice)
    {
        case 1: cout << "You chose calculate for a circle\n\n";
                calc_circle();
                break;
        case 2: cout << "You chose calculate for a rectangle\n\n";
                calc_rectangle();
                break;
        case 3: cout << "You chose calculate for a triangle\n\n";
                calc_triangle();
                break;
        case 4:
                return 1;
    }//!End of switch/case
    return 0;
}

/* . . . same code as above . . . */

---

Blue lines are changed, the actual changes are in bold.

I think your while loop does the opposite of what you want, dwks. Returning 0 will break the loop, returning 1 will continue it.

Yes, indeed. I meant to put while(!menu()) as I originally indicated.