C++ Undeclared Identifier in Template Template Parameter

Question

I have a C++ class that is templatized like so:

template <typename Operator>
class MyClass;

Where Operator can also be templatized itself into:

template <typename Param1, typename Param2, typename Param3>
class MyOperator;

Now, when I try to write a templatized method for the class, MyClass, I get an error - this code:

template < template < typename Param1, typename Param2, typename Param3 > typename Operator >
void MyClass<Operator<Param1, Param2, Param3>>::FunctionName()

Produces the error: "undeclared identifier" for each of Param1, Param2, Param3 and Operator. Why would this be, since the typenames/classes are specified right above?

I know the example code doesn't make that much sense, but my ultimate goal is to partially specialize it to look something like:

template < template < typename Param1, typename Param2, typename Param3 > typename Operator >
void MyClass<Operator<Param1, "CustomParam", Param3>>::FunctionName()

So that if the second Param is "CustomParam", the function would execute a specific implementation. Would this work, even though I specify all the parameters as template parameters (since the parameter I want to specialize is the second parameter, but the first is not specialized)? Thanks!

please include a [mcve] – 463035818_is_not_an_ai May 13 '20 at 07:29 — 463035818_is_not_an_ai, May 13 '20 at 07:29

Jarod42 · Accepted Answer · 2020-05-13T08:49:02.887

Parameter names in template template parameter are just informative (as names for parameters in function pointer void (*f)(int a, int b) (a and b cannot be used)), you should do:

template <template <typename, typename, typename> typename Operator,
          typename Param1, typename Param2, typename Param3>
void MyClass<Operator<Param1, Param2, Param3>>::FunctionName() {/*...*/}

Notice that you cannot partial specialize method/function, you have to partial specialize the whole class.

C++ Undeclared Identifier in Template Template Parameter

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