CODEWITHSUNDEEP
c++templatesfunctionparameters
Yola
Yola

Reputation: 19093

how to pass function pointer as a template parameter

second line compiles well, but third gives me an errors

int (* const f)() = ff;
cout << typeid(replace<int (*)(), int, char>::type).name() << endl;
cout << typeid(replace<f, int, char>::type).name() << endl;

test.cpp:3:25: error: the value of ‘f’ is not usable in a constant expression 
test.cpp:1:15: note: ‘f’ was not declared ‘constexpr’
test.cpp:3:37: error: template argument 1 is invalid

Upvotes: 1

Views: 968

Answers (2)

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 507393

A template argument can be a function pointer only if the parameter is a non-type parameter. But sadly C++11 does even not yet allow to use the full variety of constant expressions to compute a template argument. For a non-type pointer or reference, you are only allowed to pass the value of another template parameter or the directly obtained address of the function or object, without storing it first in a const / constexpr variable.

This limitation will most probably be lifted for next C++ revision.

Upvotes: 3

AlexTheo
AlexTheo

Reputation: 4184

You could use templated interface instead of function pointer:

template<class T>
class IMyInterface{
public:
 virtual void doSomething(const T& arg) =0;
};

Upvotes: 0

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