Defining and calling a C++ function of a specialized template

Question

I am currently learning C++ in-depth, and I have come across something that has stumped for a couple hours now. Why is it when I make a template and then specialize it, that I can't call or define that function for the specialized version? The compiler complains, and I have scoured Google for a possible hint as to what I am doing wrong, but to no avail. I am very sure it is something very simple that I am overlooking:

template <typename T>
class C { };

//specialization to type char
template <>
class C <char> 
{
  public:
    void echo();
};

//compiler complains here
template <>
void C <char> :: echo() 
{
  cout << "HERE" << endl;
}

error: template-id ‘echo<>’ for ‘void C::echo()’ does not match any template declaration

Demo.

possible duplicate of [template-id does not match any template delcaration](http://stackoverflow.com/questions/4694181/template-id-does-not-match-any-template-delcaration) — GWW, Oct 28 '11 at 04:49

score 7 · Accepted Answer · edited Jun 14 '23 at 16:51

7

//specialization to type char
template <>
class C <char>
{
  public:
    void echo();
};

//template<>  <----- don't mention template<> here
void C <char> :: echo()
{
  cout << "HERE\n";
}

P.s. Never say endl when you mean '\n'. What is the C++ iostream endl fiasco?

edited Jun 14 '23 at 16:51

Hari

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answered Oct 28 '11 at 04:47

Robᵩ

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Thanks @Rob! Last question, why does the function "echo()" not need a template specifier? – jrd1 Oct 28 '11 at 05:02
2

@jrd1, because `echo()` is not a `template` function by itself. It's a member method of a `template class`. – iammilind Oct 28 '11 at 05:08

score 0 · Answer 2 · answered Jun 14 '23 at 06:32

Robᵩ's answer fixes the issue. It will be good to know more about the reason.

From Explicit (full) template specialization:

Members of specializations

When defining a member of an explicitly specialized class template outside the body of the class, the syntax template<> is not used, except if it's a member of an explicitly specialized member class template, which is specialized as a class template, because otherwise, the syntax would require such definition to begin with template required by the nested template.

Defining and calling a C++ function of a specialized template

2 Answers2