How can I do a template function that returns vector of templatetype?

[Rob]

I want to do something like this:

template <typename MyType>
std::vector<MYTypeget(SomeClass c)
{
...elided...
}
 
int main()
{
std::vector<std::vector<std::stringret = get( c );
}
 

But the problem is it has no way (I can see) to differentiate based on
type. How do I do something like that?

[Davis King]

On Apr 11, 9:38 pm, Rob <[e-mail]:

I want to do something like this:

template <typename MyType>
std::vector<MYTypeget(SomeClass c)
{
     ...elided...

}

int main()
{
    std::vector<std::vector<std::stringret = get( c );}

 

But the problem is it has no way (I can see) to differentiate based on
type. How do I do something like that?

In your example you would say:
std::vector<std::vector<std::string ret =
get<std::vector<std::string( c );

That is, you would explicitly provide the template argument at the
point you call get() since in your case it isn't inferable from the
argument to the function. Although maybe this isn't what you are
asking?

-Davis

[Rob]

On Apr 11, 9:55*pm, Davis King <[e-mail]:

On Apr 11, 9:38 pm, Rob <[e-mail]:

I want to do something like this:

 template <typename MyType>
 std::vector<MYTypeget(SomeClass c)
 {
 * * *...elided...

        
                 }
 

 

        

                 int main()

 {

 * * std::vector<std::vector<std::stringret = get( c );}