长度未知的std :: array作为模板功能规范的类型

Question

I want to pass std::array class with a known type, but unknown size to the template function specialization:

class foo {
public:
void bar<class T>(T arg1, int arg2, int arg3) {
    //For any other T
}
void bar<std::array<rs::float3, size>>(T arg1, arg2) { 
    //Specific function for type std::array of rs::float3's and unknown length (size is not a type, just variable).
}

Answer 1

What about something like

template <typename T>
void foo (T const &)
 { std::cout << "foo generic" << std::endl; }

template <std::size_t Dim>
void foo (std::array<float, Dim> const &)
 { std::cout << "foo float array dim " << Dim << std::endl; }

?

The following is a full working example

#include <array>
#include <iostream>

template <typename T>
void foo (T const &)
 { std::cout << "foo generic" << std::endl; }

template <std::size_t Dim>
void foo (std::array<float, Dim> const &)
 { std::cout << "foo floar array dim " << Dim << std::endl; }

int main ()
 {
   foo(0);
   foo(std::array<float, 12U>{}); 
 }

Answer 2

If you want the function to accept array objects of different sizes, you pretty much need to make it a function template, and pass the size as a template parameter:

template<size_t N>
void bar(std::array<rs::float3, N> const &arg1, 
         std::array<rs::float3, N> cons t&arg2) 
{
   // ...
}

In this case, I guess you also want it to be a partial specialization of another template. Unfortunately, there is no partial specialization of function templates. There is a part of the standard that deals with partial ordering of function templates, but the usual recommendation is that you just use overloading instead.

Answer 3

Try this:

class foo
{
public:
    template <class T>
    void bar(T arg1, int arg2, int arg3) {
        //For any other T
    }

    template <size_t size>
    void bar(std::array<rs::float3, size> arg1, int arg2) { 
        //Specific function for type std::array of rs::float3's of caller-specified length
    }
};

Live demo