Pointers to objects -- passing and returning pointers

This is a discussion on Pointers to objects -- passing and returning pointers within the C++ Programming forums, part of the General Programming Boards category; I'm writing a large numbers class just for fun and currently use no pointers. I basically have objects with functions ...

  1. #1
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    Pointers to objects -- passing and returning pointers

    I'm writing a large numbers class just for fun and currently use no pointers. I basically have objects with functions for performing operations and these functions return the answer as another LargeNumber. So
    Code:
    (LargeNumber)c = (LargeNumber)a.add((LargeNumber)b);
    will return another (LargeNumber)c that equals a + b. What I want to do is define it to take and return pointers because this would be (supposedly) faster. I was working with it earlier and got a couple errors about scope and pointers so I figure I was doing it wrong.

    Here's my multiply function, it's short because it uses add() and other routines, I added a couple comments.
    Code:
    LargeNumber LargeNumber::mul(LargeNumber x)
    {
      LargeNumber muls[2] = LargeNumber(0,0,0);
      muls[0] = simplemul(x.getDigit(x.getLength()));//simplemul multiples the entire number by a single digit
      muls[0].setSign(1);//sign figured out latter, all positive for addition
      bool yy = 1;
      for(int i = x.getLength() - 1; i > 0; i--)//elementary arithmetic, multiply top by each digit in the bottom...
      {
        muls[yy] = simplemul(x.getDigit(i));
        muls[yy].setSign(1);
        muls[yy].mul10(x.getLength() - i); //adds a zero to the right
        muls[yy] = muls[yy].add(muls[!yy]);
        yy = !yy;
      }
      muls[!yy].setDec(getDec() + x.getDec());//finds the decimal
      if(getSign() == 0 && x.getSign() == 0)
        muls[!yy].setSign(1);
      else
        muls[!yy].setSign(getSign() && x.getSign());
      return muls[!yy];
    }
    I want to turn this into a function that takes a pointer of a LargeNumber and returns a pointer to another LargeNumber.


    Just for simplicity I don't need you to rewrite the entire function. Just pointify this and/or explain something
    Code:
    LargeNumber* LargeNumber::mul(LargeNumber* x)
    {
      LargeNumber muls[2] = LargeNumber(0,0,0); //Define muls[2] somehow...
      muls[0] = simplemul(x.getDigit(x.getLength()));
      //do stuff (I understand x->function()??)
      return muls[yy];
    }
    Code:
    LargeNumber* a = &LargeNumber(...);
    LargeNumber* b = &LargeNumber(...)
    LargeNumber* c = a->mul(b);
    
    //Alternatively I assume this would work??
    
    LargeNumber a = LargeNumber(...);
    LargeNumber b = LargeNumber(...);
    LargeNumber* c = a.mul(&b);
    Last edited by 1veedo; 04-04-2008 at 10:09 AM.
    --

  2. #2
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    Code:
      if(getSign() == 0 && x.getSign() == 0)
        muls[!yy].setSign(1);
      else
        muls[!yy].setSign(getSign() && x.getSign());
    will be more correctly as:
    Code:
        muls[!yy].setSign(getSign() != x.getSign());
    That sets sign to 1 if one of the signs is non-zero but the other one is zero.

    Also, if you are going to use pointers, you are probably better off with a three-way-operator function (where one operand is "this", of course), such as this:
    Code:
    void LargeNumber::mul(LargeNumber *x, LargeNumber *res)
    {
    ...
    }
    Just beware that "res" could be the same variable as y or this.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  3. #3
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    What's the "proper" way to do this? I've seen code in other projects that make heavy use of pointers to objects. I figure the way I'm currently doing it is inefficient because two object are being copied every time I call a function. You're recommending something like this?
    Code:
    LargeNumber a = LargeNumber(...);
    LargeNumber b = LargeNumber(...);
    LargeNumber c = LargeNumber(...);
    a.mul(&b, &c); //give c the value of a times b
    This isn't the way I always see it being done... The functions of course could somehow be static (like you can do in Java) or not attached to the objects so you could do
    Code:
    LargeNumber.mul(&a, &b, &c); //multiply a and b, give c the value.
    Btw how does new work? http://en.wikipedia.org/wiki/New_(C%2B%2B)
    I'm trying to get a little more in depth with C++.

  4. #4
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    I figure the way I'm currently doing it is inefficient because two object are being copied every time I call a function.
    Generally, you can avoid copying by passing by (const) reference. You could then do something like:
    Code:
    LargeNumber LargeNumber::mul(const LargeNumber& rhs) const;
    // ...
    LargeNumber a(...);
    LargeNumber b(...);
    LargeNumber c = a.mul(b);
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  5. #5
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    I would make mul a free function and use it like this (you could implement it directly or use an idea that follows a common operator+ idiom):
    Code:
    void LargeNumber::multiply_by(const LargeNumber& rhs)
    {
      // implementation ...
    }
    
    LargeNumber multiply(const LargeNumber & lhs, const LargeNumber& rhs)
    {
        LargeNumber temp(lhs);
        temp.multiply_by(rhs);
        return temp;
    }
    
    ...
    
    LargeNumber a(...);
    LargeNumber b(...);
    LargeNumber c = multiply(a, b);
    Note that the return value of your mul function should not be a pointer or reference, because by multiplying two objects you are creating a third object that is the result. You use pointers and references only to objects that already exist.

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