Question

• I have a template class A, and a stream operator function for A that is consequently a template function.
• I have a class B that inherit from A, specifying a type for the template and adding some stuff specific to B.
• I am writing a stream operator on B that does something specific to B and then falls back to the rest of the stream operations programmed for A.

Because B is a subclass of A (do you call it that?), invoking A’s stream operator should work. In fact, operator>>(f,a) in main() works. But for some reason it doesn’t work on b casted to A. The error I get is “no matching function for call to ‘operator>>(std::basic_istream >&, A)”.

What am I doing wrong?

Here is the sample code:

#include <stdlib.h>
#include <fstream>
#include <iostream>

using namespace std;


template <class TypeT>
class A
{
public:
    A(){a = -9;}
    A(TypeT v){a = v;}
    TypeT a;
};

class B : public A<int>
{
public:
    B(int w) : A<int>(10) {b = w;}
    int b;
};

template <class TypeT>
istream &operator>> (istream &s, A<TypeT> &a)
{
    cout << "a.a = " << a.a << endl;
    return s;
}

istream &operator>> (istream &s, B &b)
{
    cout << "b.b = " << b.b << "  ";
    operator>>( s, (A<int>)b);    // error!
    return s;
}



int main(void) {
    ifstream f("/dev/null");

    A<int> a(0);
    operator>>( f, a );

    B b(1);
    operator>>( f, b );

    return EXIT_SUCCESS;
}

.

Answer 1

Change your cast to:

operator>>(s, (A<int> &)b);

Your original cast creates a temporary which you can only get a const reference to.