How to specify the template parameters of a template member of a template class?

[JUANDENT]

Consider the following:

template<typename T> class Blob
{
public:
	template<typename It>
	Blob(It b, It e) : data(std::make_shared<std::vector<T>>(b,e)) {}


private:
	std::shared_ptr<std::vector<T>> data;
};

What if It cannot be deduced from the constructors arguments? How do you specify it?

using T = std::vector<int>::iterator;
Blob<int> blob<T>(v.begin(), v.end());

??
this does not compile!

[JLBorges]

It does compile cleanly. http://coliru.stacked-crooked.com/a/9d302776493a094a

[JUANDENT]

I know the first code compiles.

my question is the following: assume the template parameter It of the constructor wasn't deduced, how could one specify the ctor template??


what does not compile is

using T = std::vector<int>::iterator;
Blob<int> blob<T>(v.begin(), v.end());

I know in this case the template of the ctor is deduced and there is no problem. My question is ASSUME It WAS NOT deduced, then HOW can we specify it?

[mbozzi]

If the question is "how do I explicitly specify a constructor's template arguments?" the unfortunate answer is "you can't".
[temp.arg.explicit] has this note:

[Note 4: Because the explicit template argument list follows the function template name, and because constructor templates ([class.ctor]) are named without using a function name ([class.qual]), there is no way to provide an explicit template argument list for these function templates.
— end note]

https://eel.is/c++draft/temp#arg.explicit-8

[JUANDENT]

that's what I was fearing!!

Thank you!!