Thread: Help with Homework... Time is almost up.

actually, he was right... the 'n' they were referring to was here:

Code:

int input[n]; //there's the n

input[0]; //there's the 0 (first element)
...
input[n-1]; //there's the n-1 (last element)

So, input[2] means actually 2 elements - 0 and 1 (n-1).

Your Welcome.

Originally posted by major_small
I dont' do flowcharts or trace tables or anything like that before programming... only for really big projects... and don't mix programming and building a house... that is one of the worst analogies i've ever heard...

It's a perfect analogy. You need to have a plan when you build a house or a program. It's just that many programs are more on the order of building a doghouse instead of a mansion -- the blueprints are simple enough that experience makes them unnessesary.

But inexperiance make them very useful.

I don't know about that... I never used them for small stuff, even when I was inexperienced...

>>because input[2] means only two elements not three in the array but u needed here three elements of the array input

by my counting input[2] has three elements: [0].[1].[2] !!

Ok, here is the current work (sorry about the formatting... don't know html tags to display it properly on this forum). I have few issues still, but overall I am pleased ( this is my first Computer Science class). My main problems still are the following:

1. Getting the program to immediately give an error if the input
given is not a-f instead of it asking for the integers first.

2. For case 'c', I need to find the average of the 3 integers
converted to the floating point average. The compiler
warns me that it is coverting it with a possible loss of
data. Am I doing this right or is there another way?

3. The inability to handle a zero input for case 'a' is still
present as I focused all my energy I finishing the rest
of the program first. I will be working on that in the
meantime.

4. Anything else that you feel like mentioning.

Code:

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

int main()
{
   int val1;
   int val2;
   int val3;
   int first;
   int second;
   int third;
   char option;
   float sum;
   int same;
   int even = 0;
   int odd = 0;
   float num1 = 0;
   float num2 = 0;
   float num3 = 0;

// Title, Input

   cout << "Integer Calculation Program\n" << endl;
   cout << "Enter a single character a-f\n" << endl;
   cin >> option;
   cout << "Enter three integer values." << endl;
   cin >> val1 >> val2 >> val3;
  
  
   switch (option)

/* 
   Case 'a' compares the 3 integers, assigns the values in order from smallest
   to largest to the variables first, second, and third, and then displays
   the output to the user.
*/

   {
   case 'a'  : 
     
     if (val1 <= val2 && val1 <= val3 && val2 <= val3)
	 
                 {   
                   first = val1;
                   second = val2;
                   third = val3;
	 }
   
     if (val2 <= val1 && val2 <= val3 && val1 <= val3)

	 {
                   first = val2;    
                   second = val1;
                   third = val3;
	 }
   
   
     if (val3 <= val1 && val3 <= val2 && val2 <= val1)

	 {
                   first = val3;
                   second = val2;
                   third = val1;
	 }   
   
     if ( val3 <= val1 && val3 <= val2 && val1 <= val2)

	 {
                   first = val3;
                   second = val1;
                   third = val2;
	 }
   
    cout << first << " " << second << " " << third << " " << endl;
                               
	                                                       break;
/* 
   Case 'b' calculates the absolute values of the 3 integers and then
   displays the output to the user.
*/
														   
   case 'b'  :
	
       val1 = abs (val1);
       val2 = abs (val2);
       val3 = abs (val3);

       cout << val1 << ", " << val2 << ", and " << val3 
               << " are the absolute values of your input." << endl;

	                                                       break;
/*
   Case 'c' adds the 3 integers and then finds the average floating point
   value of their sum by dividing by 3.
*/

   case 'c'  :
     

     sum = (val1 + val2 + val3) / 3.0; 
     cout << sum << " is the average of the three integers given."
             << endl;

	                                                       break;

/*
   Case 'd' compares the 3 integers and determines and displays the number of 
   the values that are the same.
*/


   case 'd'  :

     if (val1 == val2 && val2 == val3)
		 same = 3;
    
     if (val1 == val2 && val3 != val1)
		 same = 2;
    
     if (val2 == val3 && val1 != val2)
		 same = 2;
    
     if (val3 == val1 && val2 != val3)
		 same = 2;
    
     if (val1 != val2 && val2 != val3 && val1 != val3)
		 same = 0;

     cout << same << " of the integer values are the same." 
             << endl;

	                                                       break;

/* 
   Case 'e' determines how many of the 3 integer values are even and how many
   are odd and then displays the information to the user.
*/

   case 'e'  :

       if ((val1 %2) == 0)
	      even++;
          else
                        odd++;
	  
       if ((val2 %2) == 0)
	      even++;
          else
                        odd++;

       if ((val3 %2) == 0)
                      even++;
          else 
	        odd++;
	
       cout << even << " of the values are even." << endl;
       cout << odd << " of the values are odd." << endl;

	                                                       break;
	     
/*
   Case 'f' determines if the 3 integers are positive, and then adds the 
   positive integers together and divides them by 3.0 to determine the 
   float point average of the positive integers.
*/

   case 'f'  :

	  if (val1 > 0)
		   num1 = val1;
	  if (val2 > 0)
		   num2 = val2;
	  if (val3 > 0)
		   num3 = val3;
	  sum = (num1 + num2 + num3) /3.0;
	  
	  cout << sum << " is the floating point average "
                          << "of the positive integers."  << endl;

	                                                       break;

/*
   If any character besides a-f was given, then after entering the 3 integer
   values, the program gives an error message informing the user of the
   mistake.
*/

   default : cout << "Not a valid option. You did not choose "
                         << "a character a-f." << endl;
   }




return 0;

}

>>Not sure if you were joking axon,

lets just say I was joking

hey uncertain please use [code][/code] tags whenever posting any code! please edit your most recent post with the [code][/code] tags surrounding you code. Also, reread the rules, and other sticky notes on top of this forum.

>>Sometimes the most obvious just slips right past my eyes. Thanks for the tip Axon.

np. looks much better, doesn't it?

for part 1 and 2, just have an if statement checking if the input is out of bounds...if it is display message and loop back. For 2) also an if statement if c, read in floats...pseudo code below:

Code:

float a, b, c;
bool notDone = true;

while( notDone )
{
cout << "Integer Calculation Program\n" << endl;
   cout << "Enter a single character a-f\n" << endl;
   cin >> option;

   if( option not a-f )
        cout << "error" << endl;
   else
         notDone = false;
}//while
   if( option is c)
    { cout << "Enter three floats." << endl;
   cin >> a >> b >> c; //then use these in case c
      }
     else
      {
       cout << "Enter three integer values." << endl;
       cin >> val1 >> val2 >> val3;
       }

the way you have your code before, the compiler is giving you warning because you are dividing an int by a float...

if someone inputs 3.3333 and you read this in into an int...it will only keep the 3. Some compilers will give you an error.