Friend functions with templates

Question

I have a generic class , Array1d, with a friend function declared as,

friend std::ostream& operator<< <>(std ::ostream& out, Array1D<T>& a);

and defined as

template<typename U>
std::ostream& operator<< (std ::ostream& out, Array1D<U> a){
    for(int i=0;i<a.size;i++){
        out<<a[i]<<" ";
    }
    out<<endl;
    return out;
}

but if I try,

Array1D<int> a;
cout<<a;

I get this error

(1).cpp|62|error: template-id 'operator<< <>' for 'std::ostream& operator<<(std::ostream&, Array1D<int>&)' does not match any template declaration|

I have tried instatiating it explicitly for int,

std::ostream& operator<< (std ::ostream& out, Array1D<int> a){
    for(int i=0;i<a.size;i++){
        out<<a[i]<<" ";
    }
    out<<endl;
    return out;
}

But it gives the same error. Help appreciated.

A minimal, working example to be used to reproduce the error would be appreciated too. — skypjack, Aug 27 '16 at 14:16

score 1 · Accepted Answer · answered Aug 27 '16 at 13:53

1

friend std::ostream& operator<< <>(std ::ostream& out, Array1D<T>& a);
template<typename U> std::ostream& operator<< (std ::ostream& out, Array1D<U> a)

These two are not the same functions. This is because Array1D<T> is a concrete type. If you want 1. to match for 2., you need to make it a template, possibly by making it Array1D<U>. You might as well check that T=U, if you want to be over-cautious.

answered Aug 27 '16 at 13:53

lorro

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Is there a way to do this, without defining the function inside the class? – Metafity Aug 27 '16 at 13:58
Yes, the template is still defined outside the class. – lorro Aug 27 '16 at 14:29

Friend functions with templates

1 Answers1