Re: Copy constructor not called
* Christoph Bartoschek:
the copy construcotr I provided for the following class is not [called].
Instead I think that a compiler generated one is used. Can anyone give me a
hint why this is the case?
#include <iostream>
class P {};
template <typename T> class Type {
public:
Type() { std::cerr << "Default constructor: " << this << std::endl; }
template <typename U> Type(Type<U> const & other)
{ std::cerr << "Copy constructor: " << this << std::endl; }
A templated constructor is not a copy constructor, by definition.
~Type() { std::cerr << "Destructor: " << this << std::endl; }
};
int main() {
Type<P> src;
Type<P> tgt(src);
}
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