CODEWITHSUNDEEP
c++templatestemplate-meta-programming
Parankush
Parankush

Reputation: 97

c++ meta-function returning a type

I am trying to create a meta-function that returns a type. The inspiration came from the first accepted answer in C++ meta-function over templates . I want my meta-function to return the type char and then I want to use that output to create a char variable. I believe my template works but I have no idea how to check the output. I was unable to print it. How do I check what the returned type is? How do I use that returned type to create a variable?

This is a summarized version of my code:

template<typename T> struct AA;
// allow for shorter syntax - decl<...>
template<typename T> using decl = typename AA<T>::result;

// char termination
template<ExprType eType>
struct AA<Expr<eType>> { using result = char; };

...other templates that specialize and allow for recursion

int main()
{
    decl<..many args..> typee;

    return 0;
}

Upvotes: 0

Views: 615

Answers (2)

Quentin
Quentin

Reputation: 63114

My usual method is to declare a dummy class template:

template <class...>
struct check_type;

And use it in a way that triggers an error:

check_type<decltype(typee)>{};

This way the type of whatever I passed appears in the compiler's output:

prog.cc: In function 'int main()':
prog.cc:8:18: error: invalid use of incomplete type 'struct check_type<int>'
    8 |  check_type<int>{};
      |                  ^
compilation terminated due to -Wfatal-errors.

Upvotes: 1

Jarod42
Jarod42

Reputation: 217085

You can check type with

static_assert(std::is_same_v<char, decl</*..many args..*/>>)

And for debugging, you might use

template <typename> struct Debug; /* No definition */

Debug<decl</*..many args..*/>> d; // Error similar to: No definition for Debug<char>

Upvotes: 1

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