Thread: Iterator

Okay here's the deal w/inheriting std::vector and the template arguments:

Case 1:

Code:

#include <vector>

class SizeVector : public std::vector<int>{ // template parameter is defined

};

int main(){

  SizeVector szvec; // template parameter not needed
  return 0;
}

Case 2:

Code:

#include <vector>

template <typename T>
class SizeVector : public std::vector<T>{ // template parameter ambiguous

};

int main(){

  SizeVector<int> szVec; // template parameter needed
  return 0;
}

So it depends on what you need your class to do I guess.

Here's an example of how to do this in SizeVector's initialization list:

Code:

#include <iostream>
#include <vector>

template <typename T>
class SizeVector : public std::vector<T> {
	
	public:
		
		SizeVector<T>() : std::vector<T>() {} // Default constructor
		
		SizeVector<T>( const std::vector<T>& rhs ) // Alternate constructor copies an existing std::vector
		: std::vector<T>( rhs ) {}
};

int main(){
	
	std::vector<int> aVector;
	aVector.push_back( 5 );
	
	SizeVector<int> szVector( aVector );
	
	std::cout << szVector[0] << "\n";
	
	return 0;
}

Don't inherit from std::vector.

If you want a mathematical vector ("an element of vector space"), an iterator seems unneeded to me.

I won't derive from std::vector, since that purpose does not fulfill the Liskov substitution principle?

No, in this case it's simply a language-technical issue. vector is not meant to be derived from. You can do it, but there's restrictions you'll have to observe. Anyway, the C++ way of doing things is to supply the additional functionality you want in free functions, or to write a class that has a vector as a member.

And a math vector with a variable number of dimensions (which is what one built from std::vector would be, or else there's little point in using std::vector) isn't terribly useful.