CODEWITHSUNDEEP
c++
TheLogicGuy
TheLogicGuy

Reputation: 691

How to call a templated function with a parameter in the template?

I've this function:

template <typename T, T sep>
void split (){
    std::cout << sep << std::endl;    
}

When I try to call it with this command: split<'f'>();
I get the following error:

q3.cpp: In function ‘int main()’:
q3.cpp:36:16: error: no matching function for call to ‘split()’
     split<'f'>();
                ^
q3.cpp:36:16: note: candidate is:
q3.cpp:31:6: note: template<class T, T sep> void split()
 void split (){
      ^
q3.cpp:31:6: note:   template argument deduction/substitution failed:

Why?

Upvotes: 1

Views: 52

Answers (1)

StoryTeller - Unslander Monica
StoryTeller - Unslander Monica

Reputation: 170055

Why?

Because the first template parameter is a type, not a value. 'f' is character constant, a value. And you cannot plug it in for a type.

A correct call would be split<char, 'f'>().

In the upcoming C++17 standard, you can in fact redefine your template in a way that allows the syntax you want:

template <auto sep>
void split (){
    std::cout << sep << std::endl;    
}

Now the call split<'f'>() will deduce the type of sep.

Upvotes: 8

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