问题 函数指针作为模板参数

在下面的C ++代码中，该行 bar<func_ptr>(); //does not work 导致编译错误：

#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;
}

g ++的输出是这样的：

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

当我直接传递地址时，我不明白为什么它有效 foo 作为模板参数但是当我通过constexpr时 func_ptr，代码不再编译，即使它完全包含该地址 foo 在编译时。谁可以给我解释一下这个？

编辑：我的g ++版本是

$ 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.