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
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