Overloading HugeInteger

figured out the bigNumber error, but now instead of that error my fourth error is this:
1>d:\homework\visual studio 2005\projects\olhugeinteger\olhugeinteger\olhugeinteger.cpp(101) : error C2676: binary '[' : 'const OLHugeInteger' does not define this operator or a conversion to a type acceptable to the predefined operator

jjust removed the 2 from bigNumber
ostream &operator<<(ostream &output, const OLHugeInteger &bigNumber)

ok now I did input, and I get two errors.  Here is my input>>

istream &operator>>(istream &input, OLHugeInteger &bigNumber)
{
      //characters read in to hold integer values
    char line[41];      
     
      //ask for usre input and make sure no more than 40 digits where entered
      //if more than 40 digits are entered, drop extra from bigNumber
      cout << "Enter Huge Integer (Max: 40 digits, extra digits will be dropped!): " << endl;
      cin.getline(line, sizeof line);
      cin.sync();
      cin.clear();

    //determine size of users integer
      inputSize = strlen(line);      

      //check for negative integer value and set bool
      if (inputSize > 0)      
      {
            int negative = '-';
            if (line[0] == negative)
                  bigNumber.isNegative = true;
      }

      //set negative to zero
      if (bigNumber.isNegative)
            line[0] = '0';

      //check that all digits entered where integer values
      for (int loopSize = 0; loopSize < inputSize; loopSize++)
            if (!isdigit(line[(inputSize - 1) - loopSize]))
            {
                  cout << "Not all digits entered where integers.  Array initialized to zero" << endl;
                  return;
            }

      //add digits to array bigNumber
      for (int loopSize = 0; loopSize < inputSize; loopSize++)
            //subtract ASCII value for '0' to get raw value store digits at
            //end of 40 digit array all non initialized parts of array stay a 0
            input >> bigNumber[(bigNumber.arraySize - 1) - loopSize] = line[(inputSize - 1) - loopSize] - '0';
      
      return input;
}//end input function

Here are my errors:
1>d:\homework\visual studio 2005\projects\olhugeinteger\olhugeinteger\olhugeinteger.cpp(75) : error C2561: 'operator >>' : function must return a value
1>        d:\homework\visual studio 2005\projects\olhugeinteger\olhugeinteger\olhugeinteger.h(27) : see declaration of 'operator >>'
1>d:\homework\visual studio 2005\projects\olhugeinteger\olhugeinteger\olhugeinteger.cpp(82) : error C2676: binary '[' : 'OLHugeInteger' does not define this operator or a conversion to a type acceptable to the predefined operator

The first error is the line which just has:
return;

You need to return an istream.  For instance:
return input;

The next error is on the line:
input >> bigNumber[(bigNumber.arraySize - 1) - loopSize] = line[(inputSize - 1) - loopSize] - '0';

First of all, you need something more like:
input >> bigNumber.bigNumber[(bigNumber.arraySize - 1) - loopSize] = line[(inputSize - 1) - loopSize] - '0';
(notice the bigNumber.bigNumber).

This still won't compile though.  I think it would help you to break this line into smaller chunks to help you understand what the code is doing.  For instance is this what you meant to do?
int i;
input >> i;
bigNumber.bigNumber[(bigNumber.arraySize - 1) - loopSize] = i;

I see what you mean aobut the int, i did int digit = bigNumber.bigNumber blah blah blah, but as for the first error, the return originally was in place for the error, and if I do return input, the program just keeps doing blank lines after I input an integer, for example 12, then hit enter it just goes to the next line with no print out.  I don't know if it helps but here is the part of main that pertains to it.

//input huge integer one
      OLHugeInteger one;
      cin >> one;

      //input huge integer two
      OLHugeInteger two;
      cin >> two;

as you can see it should repeat meaning ask for an integer twice, but without return input I get the above error and with return input it goes to the next line and it is blank rather then asking for another integer.  Hope this makes sense.

I'm not sure what the goal of that function is.  Do you want each integer placed in the bigNumber array?

So if the user enters "1234", do you want the bigNumber array to look like:
bigNumber[0] = 1;
bigNumber[0] = 3;
bigNumber[0] = 3;
bigNumber[0] = 4;
?

Also, you definitely want to change:
cin.getline(line, sizeof line);
cin.sync();
cin.clear();
to
input.getline(line, sizeof line);
input.sync();
input.clear();

Oops, I mean to write:
bigNumber[0] = 1;
bigNumber[2] = 2;
bigNumber[3] = 3;
bigNumber[4] = 4;

ok here is input from top to bottom.  as I said this program was completed already however, we now have to take our completed HugeInteger and overload everything.  Final assignment.

ask user for input
get line
determine size
if size is greater then 0 check for negative
if negative set bool isNegative to true
if bigNumber is negative
set line[0] to '0'
check all digits are integers now
if they are not
set all digits to 0 and tell user
add digits to array (so 1234 would be)
bigNumber[36]=1
bigNumber[37]=2
bigNumber[38]=3
bigNumber[39]=4

I need the program to tell user they did not enter all integers for example they want to type 1234 but type 12e4 or 1w34 then it is set to all zeros

Well here is how you could zero out the array:

	  //check that all digits entered where integer values
	  for (int loopSize = 0; loopSize < inputSize; loopSize++)
		  if (!isdigit(line[(inputSize - 1) - loopSize]))
		  {
			  cout << "Not all digits entered where integers.  Array initialized to zero" << endl;
			  for(int i = 0; i < bigNumber.arraySize; ++i)
				  bigNumber.bigNumber[i] = 0;
			  return input;
		  }

                                              

1:
2:
3:
4:
5:
6:
7:
8:
9:

I am still getting a blank line after i try to input 1234
I am pretty sure it has nothing to do with the whole array set to zero thing becasue when I cut it out and build and then start without debugging, I get the blank llline after i input 1234 still

so there is no confusion this is my input>>

istream &operator>>(istream &input, OLHugeInteger &bigNumber)
{
      //characters read in to hold integer values
    char line[41];      
     
      //ask for usre input and make sure no more than 40 digits where entered
      //if more than 40 digits are entered, drop extra from bigNumber
      cout << "Enter Huge Integer (Max: 40 digits, extra digits will be dropped!): " << endl;
      input.getline(line, sizeof line);
      input.sync();
      input.clear();

    //determine size of users integer
      inputSize = strlen(line);      

      //check for negative integer value and set bool
      if (inputSize > 0)      
      {
            int negative = '-';
            if (line[0] == negative)
                  bigNumber.isNegative = true;
      }

      //set negative to zero
      if (bigNumber.isNegative)
            line[0] = '0';

      //check that all digits entered where integer values
      for (int loopSize = 0; loopSize < inputSize; loopSize++)
            if (!isdigit(line[(inputSize - 1) - loopSize]))
            {
                  cout << "Not all digits entered where integers.  Array initialized to zero" << endl;
                  for (int digitIndex = 0; digitIndex < bigNumber.arraySize; digitIndex++)
                        bigNumber.bigNumber[digitIndex] = 0;
                  return input;
            }

      //add digits to array bigNumber
      for (int loopSize = 0; loopSize < inputSize; loopSize++)
      {
            int digit = bigNumber.bigNumber[(bigNumber.arraySize - 1) - loopSize];
            //subtract ASCII value for '0' to get raw value store digits at
            //end of 40 digit array all non initialized parts of array stay a 0
            bigNumber.bigNumber[(bigNumber.arraySize - 1) - loopSize] = line[(inputSize - 1) - loopSize] - '0';
            input >> digit;
      }
      return input;
}//end input function

but that will put then at the beginning of the array and not the end

Oops, I thought you wanted it at the beginning.  Well, putting the characters at the end of the array instead of the beginning should be pretty trivial for you.

ya well whenever I think that, I get all screwed up, now I gotta do add brb I am sure!

Hehe, you already have written the code to do this.

(copied from your earlier post)
bigNumber.bigNumber[(bigNumber.arraySize - 1) - loopSize] = line[(inputSize - 1) - loopSize] - '0';

wow, lots of errors here, first, does this look right from .h in public part:
OLHugeInteger operator+(const OLHugeInteger &) const;
second, here is add, I haven't changed anything yet:
OLHugeInteger operator+(const OLHugeInteger &bigNumber2) const
{
      OLHugeInteger result;
                        
      //if both bigNumbers are positive or both are negative, add together
      if (this.isNegative == bigNumber2.isNegative)
      {
            int carry = 0;      //holds carry over
            //add all digits in array
            for (int loopSize = arraySize - 1; loopSize >= 0; loopSize--)
            {
                  int temp = bigNumber[loopSize];
                  result.bigNumber[loopSize] = (temp + bigNumber2.bigNumber[loopSize] + carry) %10;      //remainder
                  carry = (temp + bigNumber2.bigNumber[loopSize] + carry) / 10;
            }//end for
            result.isNegative = isNegative;
            return result;
      }//end if
      //else subtract because neg + pos and pos + neg needs to be subtracted
      else
            return subtract(opposite(bigNumber2));
      
}//end add function

I don't really have an example to go off of, just a reference to a list of operators that can be overloaded.  But no examples in the text, so I am not sure where to go from here.

ok how would i change the add part of the following from my subtract function?

//else one is positive and one is negative
      else
            return add(opposite(bigNumber2));

Nevermind, got it, moving on!

Is it better to do the equalities in .h or do them in .cpp?  I could do this:
bool operator==(const HugeInteger &) const;      //test bn1 == bn2
      bool operator>(const HugeInteger &) const;      //test bn1 > bn2

      //test bn1 != bn2
      bool operator!=(const HugeInteger &bigNumber2) const
      {
            return !(*this == bigNumber2);
      }

      //test bn1 >= bn2
      bool operator>=(const HugeInteger &bigNumber2) const
      {
            return !(bigNumber2 > *this);
      }

      //test bn1 < bn2
      bool operator<(const HugeInteger &bigNumber2) const
      {
            return bigNumber2 > *this;
      }
      
      //test bn1 <= bn2
      bool operator<=(const HugeInteger &bigNumber2) const
      {
            return !(*this > bigNumber2);
      }
or should I do them insde the .cpp file that I have already?
bool OLHugeInteger::equalTo(const OLHugeInteger &bigNumber2) const
{
      //if one is positive and one is negative
      if (isNegative != bigNumber2.isNegative)
                  return false;
      
      for (int digitIndex = 0; digitIndex < arraySize; digitIndex++)
            //if any integers from first array does not equal integers
            //from second array, return false
            if(bigNumber[digitIndex] != bigNumber2.bigNumber[digitIndex])
                  return false;

      return true;

}//end equalTo predicate function

bool OLHugeInteger::notEqualTo(const OLHugeInteger &bigNumber2) const
{
      //if all first array integers are not equal to second array integers
      //return true
            if(!equalTo(bigNumber2))
                  return true;
      return false;

}//end notEqualTo predicate function

bool OLHugeInteger::greaterThan(const OLHugeInteger &bigNumber2) const
{
      //if equal return false
      if (equalTo(bigNumber2))
            return false;
      else
      {
            //If the two numbers are of the same sign
            if (isNegative == bigNumber2.isNegative)
            {
                  //If both are positve
                  if (!isNegative)
                  {
                        //the first 1st array int > 2nd array int return true
                        //logic( 200 is greater than 100 {bigger = greater})
                        for (int digitIndex = 0; digitIndex < arraySize; digitIndex++)
                        {
                              if (bigNumber[digitIndex] == bigNumber2.bigNumber[digitIndex])
                                    continue;
                              if (bigNumber[digitIndex] <= bigNumber2.bigNumber[digitIndex])
                                    return false;
                              else
                                    return true;
                        }
                  }
                  //else both are negative
                  else
                  {
                        // When both are negative, use the both-are-positive logic to
                        //get the job done flipflop with opposite and check again
                        if (isNegative && bigNumber2.isNegative)
                        {
                              OLHugeInteger positveThis = opposite(*this);         // Positive copy of this number
                              OLHugeInteger positveThat = opposite(bigNumber2);    // Positive copy of bigNumber2
                              if (positveThis.greaterThan(positveThat))
                                    return false;
                              else
                                    return true;
                        }
                  }
            }
            //else the two numbers are of differing signs
            else
            {
                  //if (-)*this and (+)bigNumber2
                  if (isNegative && !(bigNumber2.isNegative))
                        return false;
                  //else (+)*this and (-)bigNumber2
                  else            
                        return true;
            }
      return true;
      }
}//end greaterThan predicate function


bool OLHugeInteger::greaterThanOrEqualTo(const OLHugeInteger &bigNumber2) const
{
      //if equal to or greater than return true
      if (equalTo(bigNumber2) || greaterThan(bigNumber2))
            return true;
      return false;

}//end greaterThanOrEqualTo predicate function

bool OLHugeInteger::lessThan(const OLHugeInteger &bigNumber2) const
{
      //if not equal to and not greater than return true
      if (!equalTo(bigNumber2) && !greaterThan(bigNumber2))
            return true;
      else
            return false;
}// end lessThan function

bool OLHugeInteger::lessThanOrEqualTo(const OLHugeInteger &bigNumber2) const
{
      //if equal to or less than return true
      if (equalTo(bigNumber2) || lessThan(bigNumber2))
            return true;
      return false;

}//end lessThanOrEqualTo predicate function

of course I would have to change all the ones in the .cpp file and it would be simpler to just do them in .h I guess, so (another bad example in the book) does that mean if I do them in .h that I only do == and > in .cpp?

>> Is it better to do the equalities in .h or do them in .cpp?
That's up to you.  I usually will only put implementation in the header file if it fits on one line.  So maybe like:
bool operator!=(const HugeInteger &bigNumber2) const { return !(*this == bigNumber2); }

How are you doing on this?