Function pointers as template arguments

Question

In the following C++ code, the line bar<func_ptr>(); //does not work causes a compilation error:

#include <iostream> using namespace std; void foo(){ cout<<"Hello world"; }; template<void(*func)()> void bar(){ (*func)(); } int main() { using fun_ptr_type= void(*)(); constexpr fun_ptr_type func_ptr=&foo; bar<&foo>(); //works bar<func_ptr>(); //does not work return 0; }

The output of g++ is this:

src/main.cpp: In function ‘int main()’: src/main.cpp:19:16: error: no matching function for call to ‘bar()’ bar<func_ptr>(); //does not work ^ src/main.cpp:10:6: note: candidate: template<void (* func)()> void bar() void bar(){ ^~~ src/main.cpp:10:6: note: template argument deduction/substitution failed: src/main.cpp:19:16: error: ‘(fun_ptr_type)func_ptr’ is not a valid template argument for ty pe ‘void (*)()’ bar<func_ptr>(); //does not work ^ src/main.cpp:19:16: error: it must be the address of a function with external linkage

I do not understand why it works when I directly pass the address of foo as a template argument but when I pass the constexpr func_ptr, the code does not compile anymore even though it holds exactly that address of foo at compilation time. Can someone explain this to me?

EDIT: My g++ version is

$ g++ --version g++ (Debian 6.3.0-18+deb9u1) 6.3.0 20170516 Copyright (C) 2016 Free Software Foundation, Inc. This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

I remember some change with regardes to pointers to globals vs constexpr literals as being valid for template parameters since C++11. But I don't remember if this is one of those cases, nor how to find it. — Yakk - Adam Nevraumont
– Yakk - Adam Nevraumont, Commented Jul 24, 2018 at 15:32
Same experience as Jarod42. C++17 it works; C++14 spits out an error. — Eljay
– Eljay, Commented Jul 24, 2018 at 15:33

Raees Rajwani · Accepted Answer · 2018-07-24 16:24:31Z

From https://en.cppreference.com/w/cpp/language/template_parameters it says:

For pointers to functions, the valid arguments are pointers to functions with linkage (or constant expressions that evaluate to null pointer values). (until C++17).

Since constexpr fun_ptr_type func_ptr=&foo does not evaluate to a nullptr value at compile time, it fails if you run it with -std=c++14 or -std=c++11.

However C++17 imposes no such requirement on function pointer non type template parameters. It says:

The template argument that can be used with a non-type template parameter can be any converted constant expression of the type of the template parameter. (since C++17)

(There are some exceptions to the above but none apply to function pointers).

So the code you provide runs perfectly with the -std=c++17 option.

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