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From: Jonathan Wakely <jwakely.gcc@gmail.com>
Date: Sun, 13 Mar 2022 16:48:27 +0000
Message-ID: <CAH6eHdSM6s4cWM_PRPE04od_sZvnr+vZ-deaUfBGqeu7Q1WMow@mail.gmail.com>
Subject: Re: Not able to declare a template friend function inside a template
 class
To: Vishal Subramanyam <me@vishalsubramanyam.ml>
Cc: gcc-help <gcc-help@gcc.gnu.org>
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On Sun, 13 Mar 2022, 15:45 Vishal Subramanyam, <me@vishalsubramanyam.ml>
wrote:

> Hey,
> I'm trying to declare a function template as a friend for a class which i=
s
> already a template.
> So the situation is something like,
> Fraction.h:
>
>     template<typename T>
>     class Fraction{
>         ...
>         friend Fraction<T> operator + (Fraction<T> const &a, Fraction<T>
> const &b);
>

You don't need to say <T> here, just call the class Fraction


    }
>
>
> Fraction.cpp:
>
>     template<typename T>
>     Fraction<T> operator+(Fraction<T> const &a, Fraction<T> const &b) {
>         Fraction<T> frac(a.num * b.den + b.num * a.den, a.den * b.den);
>         return frac;
>     }
>
>
> But I'm getting the following warning:
>
>     warning: friend declaration =E2=80=98Fraction<T> operator+(const Frac=
tion<T>&,
> const Fraction<T>&)=E2=80=99 declares a non-template function
> [-Wnon-template-friend]
>
>
> Now, if I try to fix this by declaring my friend function as a template
> inside the class, following the example given in
> https://en.cppreference.com/w/cpp/language/friend
> (Section: Template Friend Operators, second example), my code would then
> look like:
> Fraction.h
>
>     template<typename T>
>     class Fraction{
>         ...
>         friend Fraction<T> operator + <>(Fraction<T> const &a, Fraction<T=
>
> const &b);
>     }
>
>
> I'm getting three related errors that say:
>
>     error: declaration of =E2=80=98operator+=E2=80=99 as non-function
>     error: expected =E2=80=98;=E2=80=99 at end of member declaration
>     error: expected unqualified-id before =E2=80=98<=E2=80=99 token
>
> I'm not sure what the issue is considering I'm following the exact syntax
> given in cppreference.com. Is this a compiler issue?


No, you are using the syntax for an explicit specialization of an existing
function template, which is what cppreference shows. But your example has
no previous declaration that can be specialized.


How do I resolve this?
>

1) Use the first syntax (without the redundant <T>) and ignore the warning
or disable it with -Wno-non-template-friend

or,

2) declare a function template which you specialize in the class:

template<typename T> class Fraction;
template<typename T> Fraction<T> operator +(Fraction<T> const &a,
Fraction<T> const &b);

    template<typename T>
    class Fraction{
        ...
        friend Fraction<T> operator + <>(Fraction<T> const &a, Fraction<T>
const &b);
    }

or,

3) define it as a template:

template<typename T>
    class Fraction{
        ...
      template<typename U>
        friend Fraction<U> operator + (Fraction<U> const &a, Fraction<U>
const &b);
    };










> Thanks,
> Vishal Subramanyam (20CS10081)
>
>