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Is a local class dependent if declared within a function template?


Is a local class dependent if declared within a function template?

By : user2948210
Date : November 14 2020, 04:48 PM
wish help you to fix your issue According to my understanding (and the current wording of the standard), C in your example is not dependent. Neither is A::Type, so the typename is not required.
There is a fundamental difference between nested classes of class templates and local classes in function templates: The latter cannot be specialized, thus any reference to a local class inside a function template is uniform. That is, in every specialization of f, C refers to the class C that is defined in this function template f. That is not the case with class templates as you can indeed explicitly specialize members on their own (as covered in [temp.expl.spec] /(1.6)):
code :
template <typename T>
class A { class C{}; };

template <>
class A<int>::C { int i; };


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Template function with dependent type parameters within template class

Template function with dependent type parameters within template class


By : user1571121
Date : March 29 2020, 07:55 AM
this will help The compiler is simply unable to deduce types from this context.
Suppose std::wstring::const_iterator is actually const wchar_t*, which is likely. In that case, how does the compiler know it should substitute std::wstring rather than any other type T with T::const_iterator being const wchar_t* (perhaps vector)?
code :
template <typename I1, typename I2>
static bool SomeOperator(const I1& p_Begin1, const I1& p_End1,
                         const I2& p_Begin2, const I2& p_End2)
  { /* stuff */ }
In C++ template function, why does dependent function call give "not declared" error?

In C++ template function, why does dependent function call give "not declared" error?


By : Akshai Shah
Date : March 29 2020, 07:55 AM
like below fixes the issue
Nobody has yet pointed out any part of the current Standard that says I can't.
code :
postfix-expression ( expression-listopt )
namespace dummies { void f(); }
template<typename T>
struct S {
  void f();
  void g() { 
    using dummies::f; // without it, it won't work
    f(T()); // with ::f, it won't work
  }
};

struct A { };
void f(A) { } // <- will find this

int main() {
  S<A> a;
  a.g();
}
template<class T> class Z {
public:
        void f() const
        {
                g(1); // g() not found in Z's context.
                      // Look again at point of instantiation
        }
};

void g(int);
void h(const Z<Horse>& x)
{
        x.f(); // error: g(int) called by g(1) does not depend
               // on template-argument ``Horse''
}
void Z<Horse>::f() { g(1); }
void h(const Z<int>& y)
{
        y.f(); // fine: g(int) called by g(1) depends
               // on template-argument ``int''
}
void Z<int>::f() { g(1); }
Friend template function declared inside template class causing undefined symbol link error

Friend template function declared inside template class causing undefined symbol link error


By : Yaniv Friedensohn
Date : March 29 2020, 07:55 AM
How to define a friend function declared in a non template class internal to a template class outside of both classes?

How to define a friend function declared in a non template class internal to a template class outside of both classes?


By : Geetha Shetty
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Here's the issue. Despite the fact that External is a template and Internal is a dependent type. The friend function is not templated itself. Odd as it may seem, it doesn't depend on the template parameter.
When you define it inline, then each specialization of the template creates the associated friend function as well. But when it's not inline you need to provide the definition of the operator for each specialization explicitly. And you can't do that with a template, since the function is not a template.
code :
std::ostream& operator<<(std::ostream& os, External<int>::Internal const& i)
{
    return os << i.internal_value;
}
class Internal {
public:
    Internal(const External& e) : internal_value{e.value} {}

private:
    std::ostream& print(std::ostream&) const;
    friend std::ostream& operator<<(std::ostream& os, Internal const& i)
    {
        return i.print(os); // Short and sweet on account of being inline
    }

    T internal_value;
};

//....

template<typename T>
std::ostream& External<T>::Internal::print(std::ostream& os) const {
   // Provided outside of the class definition, can be as verbose as you like
   return os << internal_value;
}
How to declare a function pointer to a template function whose return type is dependent on template class

How to declare a function pointer to a template function whose return type is dependent on template class


By : Hanzo Smiley
Date : March 29 2020, 07:55 AM
may help you . There are no template function pointers, if that is what you were looking for. You need to know exactly the signature of the function instantiation if you want to have a function pointer to a particular instantiation.
You can always cheat with auto:
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