Thread: Question about "this-> pointers"

I am having trouble understanding a functon that uses a ->this pointer. I also know that the function may work without using a ->this pointer, but the point is that I don't understand how the compiler understands it, or how it actually works, after reading it.

Here's the code:

Code:

#include <iostream>
using std::cout;
using std::endl;

class CBox // Class definition at global scope
{
public:
CBox(double lv = 1.0, double bv = 1.0, double hv = 1.0)	// Constructor definition
	{
	cout << endl << "Constructor called.";
	m_Length = lv; // Set values of
	m_Width = bv; // data members
	m_Height = hv;
	}

double Volume() // Function to calculate the volume of a box
	{
	return m_Length*m_Width*m_Height;
	}

int Compare(CBox xBox)	// Function to compare two boxes which returns true (1) if the first is greater than the second, and false (0) otherwise
	{
	return this->Volume() > xBox.Volume();
	}

private:
	double m_Length; // Length of a box in inches
	double m_Width; // Width of a box in inches
	double m_Height; // Height of a box in inches
};


int main()
{
CBox match(2.2, 1.1, 0.5); // Declare match box
CBox cigar(8.0, 5.0,1.0); // Declare cigar box

if(cigar.Compare(match))
cout << endl << "match is smaller than cigar";
else
cout << endl << "match is equal to or larger than cigar";
cout << endl;

return 0;
}

The compare() function that was defined in the class definition returns 1 if the statement -this->Volume() > xBox.Volume()- is true, and 0 if not.

BUT WHY?? shouldn't it need an if (xxx.volume() > xxx.volume())
return 1;
else
return 0;

??

It looks a little "magical" to me. Any help would be really appreciated. Thanks!

The compare() function that was defined in the class definition returns 1 if the statement -this->Volume() > xBox.Volume()- is true, and 0 if not.

BUT WHY?? shouldn't it need an if (xxx.volume() > xxx.volume())
return 1;
else
return 0;

??

The result of this->Volume() > xBox.Volume() is a bool. A bool, when converted into an integer, becomes 1 if it is true, and 0 if it is false. Hence, you do not need that if statement. However, perhaps you actually want to declare Compare() as returning a bool instead of an int.

By the way, you need to improve your indentation.

You're welcome

Incidentally, have you learnt about const-correctness and passing by reference yet? If so, note that Volume() and Compare() should be const member functions, and that Compare() should take another CBox by const reference.

Also, the member variables should be initialised in the constructor initialisation list, though in this case assigning to them in the constructor body as you have done is not a problem.

Quick questions though, (I actually took this code from my c++ textbook) why would the author use int instead of bool in the book? Does that serve any specific purpose?

It could be an oversight. When I see a comparison function that returns an int, it is usually to follow the strcmp() idiom: if the number returned is less than/equal to/greater than 0, then the first object is less than/equal to/greater than the second object, respectively. However, in this case Compare() merely compares for "greater than", so implementing it by overloading operator> would be better, in my opinion.

So the expression using the boolean variable: if(cigar.Compare(match) == true) is the same as (cigar.Compare(match)) using the int?

Yes, though if (x == true) is generally the same as if (x). Perhaps you meant if (cigar.Compare(match) == 1).

I did not realize that: if (x == true) is generally the same as if (x) !! Now it makes all the sense of the world. That's why he typed it: if(cigar.Compare(match)).

You've been of enormous help. Thanks SO much again!

For someone who is already used to the fact that "if (x == true)" is the same as just "if (x)"; "if (x == true)" looks like how "if ((x == true) == true)" would look to someone who hasn't yet caught onto it, if that helps put it into perspective!

Of course it makes more sense if instead of 'x', your bool is named something like 'itIsRaining'.

(Makes for even nicer reading if you're using a Pascal-like language that has a 'then' at the end of the 'if')
E.g. "if it_is_raining then use_your_umbrella;"