Could not deduce template argument for overloaded function

Question

I've got a simple function called function for example with the following signature:

class Type { /* ... */ };

bool function( const Type& aValue ) { /* ... */ return true; }

I've got some other classes and I wanted to overload the mentioned function so that only the class derived from Base could use it:

class Base { /* ... */ };
class Derived : public Base { /* ... */ };

template < typename T >
bool function( const typename std::enable_if< std::is_base_of< Base, T >::value, T >::type& aValue ) { /* ... */ return true; }

It is working fine if I use it like this:

Derived object;
function< Derived >( object );

but if I leave the template argument I get the mentioned error (could not deduce template argument):

Derived object;
function( object ); // Compilation error (C2664).

Is there any solution where I can leave the template argument ?

(MSVC 2012)

Isnt T in non deduced context here: `std::is_base_of< Base, T >::value` ? See here for reference http://stackoverflow.com/questions/25245453/what-is-a-nondeduced-context — marcinj, May 24 '16 at 12:30
This isn't a compiler issue. T cannot be deduced from this, and therefore no compiler will succeed — Aaron McDaid, May 24 '16 at 12:30

Piotr Skotnicki · Accepted Answer · 2016-05-24T12:43:23.543

4

By introducing a nested name specifier (<T>::) you inhibit template type deduction on T.

That said, you have to let the argument expression type to be deduced by putting the enable_if elsewhere, e.g., in a return type syntax:

template <typename T>
auto function(const T& aValue)
    -> typename std::enable_if<std::is_base_of<Base, T>::value, bool>::type
{
    return true;
}

or in a template type parameter list:

template <typename T, typename = typename std::enable_if<std::is_base_of<Base, T>::value>::type>
bool function(const T& aValue)
{
    return true;
}

edited May 24 '16 at 12:43

answered May 24 '16 at 12:30

Piotr Skotnicki

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With the second solution I got this error: `error C4519: default template arguments are only allowed on a class template` for the function definition. And this for the function call: `error C2664: 'function' : cannot convert parameter 1 from 'Derived' to 'const Type &'` – p.i.g. May 24 '16 at 12:58
@p.i.g. MSVC you use doesn't support this C++11 feature – Piotr Skotnicki May 24 '16 at 13:00
@p.i.g. use the first syntax? – Piotr Skotnicki May 24 '16 at 13:06
Unfortunatelly that seems not a good solution when the user not uses the the return value. – p.i.g. May 24 '16 at 13:12
1

@p.i.g. why do you think so ? it is substituted by `bool` – Piotr Skotnicki May 24 '16 at 13:14

Could not deduce template argument for overloaded function

1 Answers1