Thread: Creating custom class for vector of pairs

I have the following code:

Code:

#include <iostream>
#include <utility>
using namespace std;

template<typename T>
class Vector {
	public:
		Vector(unsigned int length = 10) {
		}
};

template<typename P, typename Q>
class Pair {

};

template<typename P, typename Q>
class VectorOfPairs : public Vector<Pair<P, Q> > {
	public:
		VectorOfPairs(unsigned int length = 10) {
			Vector<Pair<P, Q> >::Vector(length);
		}
};

int main() {
	VectorOfPairs<int, int> v;
	return 0;
}

What I'm trying to do inside the "VectorOfPairs(unsigned int)" constructor is what we would write in Java as "super(length)". But for some reason, it gives me a compilation error:

Code:

test.cpp: In constructor `VectorOfPairs<P, Q>::VectorOfPairs(unsigned int) [with P = int, Q = int]':
test.cpp:26:   instantiated from here
test.cpp:21: error: dependent-name ` Vector<Pair<P, Q> >::Vector' is parsed as a non-type, but instantiation yields a type
test.cpp:21: note: say `typename  Vector<Pair<P, Q> >::Vector' if a type is meant

It seems to treat ` Vector<Pair<P, Q> >::Vector' as a type. But in reality, the type is ` Vector<Pair<P, Q> >', while the '::Vector' part is the constructor.

So my question is: is there anyway to tell the compiler that Vector<Pair<P,Q> > is the type?

Originally Posted by EVOEx

You can't call a constructor like:
Try:

Code:

template<typename P, typename Q>
class VectorOfPairs : public Vector<Pair<P, Q> > {
	public:
		VectorOfPairs(unsigned int length = 10) : Vector(length) {
		}
};

Actually, that would be "Vector<Pair<P, Q> >(length)" instead of "Vector(length)", apparently, since "Vector(length)" gives me a compilation error.

Either way, thank you. But I'm curious: is there really no way to call the Vector constructor actually inside the VectorOfPairs constructor?

Originally Posted by EVOEx

But why don't you just use std::vector< std:air<int, int> > ?

Because this is an assignment where we're supposed to make our own Vector and Pair classes. Sorry for not mentioning that earlier.