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Function Template Specialization with Forward Declared Type


Function Template Specialization with Forward Declared Type

By : user2185058
Date : November 20 2020, 04:01 AM
like below fixes the issue The simple workaround is to simply put the forward declaration on its own line:
code :
class SpecialT;
template<>
SpecialT* Foo<SpecialT>();


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Use a forward-declared class in a virtual function in a template baseclass where the constructor only needs the forward

Use a forward-declared class in a virtual function in a template baseclass where the constructor only needs the forward


By : Mars
Date : March 29 2020, 07:55 AM
may help you . First, as many have stated, the standard fully allows for compilers to instantiate (or chose not to instantiate) to their hearts content in this situation; by a strict interpretation of the standard, this is a code issue, not an MSVC bug. The behavior you are depending on is platform specific and therefore non-standard; while this is not a "bug" per se, it's distinctly non-portable.
However, it's interesting to understand why GCC and Clang differ from MSVC, which has to do with how vtables are set up in each compiler.
code :
class A : public B
Is it possible to use decltype to figure out the return type of a member function of a forward declared template class?

Is it possible to use decltype to figure out the return type of a member function of a forward declared template class?


By : Paradoxx
Date : March 29 2020, 07:55 AM
I wish this help you Forward declared means that you know that the class (or class template) exists, not what it contains. From this, it's obviously impossible to get information about any members, including member functions.
Can forward declared type templates participate in template specialization ?

Can forward declared type templates participate in template specialization ?


By : junr03
Date : March 29 2020, 07:55 AM
this one helps. Forward-declaring of a standard library container is undefined behavior (if you manage to compile it), so for std::vector you have to #include before defining is_vector.
For your own types you can do it though:
code :
// forward declaration of a type template
namespace my {
    template<class T, class Alloc> class vector; 
}

template<class T>
struct is_vector : std::false_type { };

// using the forward declared type template
template<class T, class Alloc>
struct is_vector<my::vector<T, Alloc>> : std::true_type { };

namespace my {
    template<class T, class Alloc = std::allocator<T>> class vector {}; 
}
How do I forward declare a template type that has been forward declared elsewhere with a defaulting

How do I forward declare a template type that has been forward declared elsewhere with a defaulting


By : user7730720
Date : March 29 2020, 07:55 AM
To fix the issue you can do The short answer is that you can't. You either have to guarantee somehow that the other declaration will never be included so you can specify your own default, or you need to depend on that other declaration.
But I get the feeling this is an XY problem, for the following statement:
Do template parameters need to be forward declared or does the type definition to be known

Do template parameters need to be forward declared or does the type definition to be known


By : dilan ranaweera
Date : March 29 2020, 07:55 AM
this one helps. All standard containers require the type to be complete. Since you're creating a vector of pointers, that's not an issue (pointer type derived from an incomplete type can be used as a type argument there). The error in your code is = NULL; part (well, unless std::vector itself is unavailable, then you additionally have to include ), which makes little sense anyway. You initialise data members in the constructor, not the class definition body. And you definitely don't need to have a pointer to the vector.
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