CODEWITHSUNDEEP
c++templatesaliasusingvariadic
Frosty
Frosty

Reputation: 33

Alias to variadic template with no arguments

I found some similar topics, but none seem to give straight answer about my particular problem.

I have a class (let's call it Foo) within some library that we are using across few modules. Now I need to transform this class to be a variadic template(so now I have: template<class... T> Foo() {}), as some new uses need exactly the same general mechanics, but need to accept additional parameters inside. Old implementations already in use do not need this.

So I transformed this into variadic template stored in one *.h file, but now wherever we already use this class I would need to change every occurrence of Foo() to Foo<>(), in order for existing code to work as it did. Now, I am not sure if there are some modules that I am not even aware of that would use this library, so I thought of a flexible solution like alias to Foo<>, that would not require those other modules to be updated after this change in library.

Is it possible in this particular example to use something like: using Foo = Foo<>, so that all occurrences of Foo that already exist in the code will automatically be treated as Foo<> without need of any changes in source code?

Would it be possible to have such alias defined in the same file as template definition/declaration, even though it would have exactly the same name as the class itself?

Upvotes: 3

Views: 93

Answers (1)

Mestkon
Mestkon

Reputation: 4046

If you rename the templatized definition of Foo to some other name FooExtended, then you can typedef the empty template to Foo

//old
class Foo;

//new
template<class... Args>
class FooExtended;

using Foo = FooExtended<>;

Upvotes: 3

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