/* An example on the use of functions :
          function with a single return       file: 5ex1.cpp
   FALL 1998
   ___________________________________
   Jacob Y. Kazakia   jyk0
   October 4, 1998
   Example 1 of week 5
   Recitation Instructor: J.Y.Kazakia
   Recitation Section  01
   ___________________________________
 
Purpose: This program uses a function  named volume to
         calculate and also output the various partial volumes
         of a composite solid body. The body consists of a
         cylinder of radius R and height hcyl and a cone of
         height hcon symmetrically sitting on top of the
         cylinder and having a base of the same radius as
         the cylinder.
 
  The function has the following arguments:
 a)  The radius of the base  ( in meters )
 b)  The height hcyl  ( in meters )
 c)  The height hcon  ( in meters )
 d)  an integer choice which takes the values 1, 2, 3
     for 1, the function returns the volume of the cylindrical
            part only.
     for 2, the function returns the volume of the conical
            part only.
     for 3, the function returns the total volume.
 
 
   Algorithm:   The volume of the cylinder is calculated by
 
                 vcyl = PI * radius^2 * hcyl
 
                The volume of the cone is calculated by
 
                 vcon = (1/3 ) PI * radius^2 * hcon
 
  Here PI is a constant set equal to 3.14159265358979
 
                */
 
#include <iostream.h>

// Prototype the function
 
 
void volume ( int choice, float radius, float hcyl, 
               float hcon );

Some programmers prefer to insert the entire definition of a function before the point that it will be used. However that approach may create confusion especially when many functions are used (typical program).

Note that this function will NOT return a value using its name volume.  Consequently  it is declared as void.  The function has four parameters, three of them float ( radius, hcyl, hcon) and one of type integer ( choice).  We should point out that the actual names of the parameters are unimportant (in fact, sometimes are excluded from the prototype) what matters is the types of the parameters, their number, and their ordering. In other words the above prototype tells our main function that the function volume has four parameters, the first being an integer and the other three floats.  Of course the programmer of main must know that the second parameter corresponds to the radius of the cylinder, the third to the height of the cylinder and the fourth to the height of the cone, however he/she can choose to give arguments to these parameters that have names of his/her choice.

//  Define the constant pi
 
const float PI = 3.14159265358979 ;
 
void main()
{
 
// declare the variables of the main function
 
int m ;    // integer to be used for choosing mode
float r ;  // the radius
float h1 ; // the height of the cylinder
float h2 ; // the height of the cone
float v_cyl, v_con, v_total ;  // the volumes
 
//  "hard code" the variables
 
r  = 2.34 ; // meters
h1 = 1.12 ; // meters
h2 = 3.12 ; // meters

// calculate volumes   by invoking the function
 
volume(1, r, h1, h2);
 
volume(2, r, h1, h2);
 
volume(3, r, h1, h2);

 
// hold the screen
 
cout<<" \n\n enter e (exit) to terminate the program....";
char hold;
cin>>hold;
}

//  definition of function volume
 
void volume( int choice, float radius, float hcyl, 
             float hcon)
{
// declare the local variables
 
float v; // variable used to store locally the calculated volume
 
if ( choice == 1)
       {
     v = PI * radius * radius * hcyl;
     cout<<" \n\n volume of cylindrical part = " << v <<" cubic meters"<< endl;
    }
else if(choice == 2)
   {
    v = PI * radius * radius * hcon / 3.;
    cout<<" \n\n   volume of conical part = " << v <<" cubic meters"<< endl;
    }
else if(choice == 3)
   {
    v = PI * radius * radius * ( hcyl + hcon / 3. );
    cout<<" \n\n        total volume = " << v <<" cubic meters"<< endl;
    }
else
   { v = 0;
     cout<<" You entered a non valid choice. The value zero will be returned";
    }
 
// no returns here
}

/*   HERE IS THE OUTPUT
 
 
 
 volume of cylindrical part = 19.2664 cubic meters
 
 
   volume of conical part = 17.8902 cubic meters
 
 
        total volume = 37.1565 cubic meters
 
 
 enter e (exit) to terminate the program....
 
 
*/