C++ : &(std::cout) as template argument

Question

Why isn't it possible to pass std::cout's address as template argument? Or if it is possible then how?

Here is what I tried:

#include <iostream>

template<std::ostream* stream>
class MyClass
{
public:
    void disp(void)
        { (*stream) << "hello"; }
};

int main(void)
{
    MyClass<&(std::cout)> MyObj;
    MyObj.disp();

    return 0;
}

And the error message I got from clang++ -std=c++11 :

main.cpp:15:11: error: non-type template argument does not refer to any declaration
        MyClass<&(std::cout)> MyObj;
                 ^~~~~~~~~~~
main.cpp:6:24: note: template parameter is declared here
template<std::ostream* stream>
                       ^
1 error generated.

and from g++ -std=c++11 :

main.cpp: In function ‘int main()’:
main.cpp:15:22: error: template argument 1 is invalid
  MyClass<&(std::cout)> MyObj;
                      ^
main.cpp:15:29: error: invalid type in declaration before ‘;’ token
  MyClass<&(std::cout)> MyObj;
                             ^
main.cpp:16:8: error: request for member ‘disp’ in ‘MyObj’, which is of     non-class type ‘int’
  MyObj.disp();
        ^

Any ideas?

Nf

Why did you parenthesize it? Also fix the access level. – Encephalogram

Try it without the parentheses: MyClass<&std::cout> MyObj; – Augmentative

I'd like to see an explanation on how this is parsed, though. – Encephalogram

Indeed: it works without the parenthesis! But Why? I tend to put parenthesis quite often especially in cases like &object.subobject where it doesn't seem clear to me whether the ampersand is going to apply to object or to subobject (although I'm sure the standard must lift that ambiguity)… I do agree however that there was no such ambiguity in the present case… – Nf

@GLorieul: () makes it an expression that requires runtime evaluation (more or less). Just throwing () everywhere without understanding what it does is a bad idea! – Mallet

@LightnessRacesinOrbit oh oh, now I see… Would you have a link to a reference that explains that in more detail ? – Nf

I don't think that's a dup. The Standard has language explicitly saying &X::m is required for pointers to member, but I can't find anything that applies here. – Augmentative

@Augmentative I at least think these Q&A are very closely related. The dup contains a standard cite however. Vote to reopen if you think this questin needs to be kept standalone. – Devoid

Following @πάντα ῥεῖ 's link, I found a nice answer – Nf

Answer 1

This fixes your code, omit the parenthesis:

#include <iostream>

template<std::ostream* stream>
class MyClass
{
public:
    void disp(void) { 
        (*stream) << "hello"; 
    }
};

int main(void)
{
    MyClass<&std::cout> MyObj;
    MyObj.disp();

    return 0;
}

Live Demo

---

A more detailed explanation why can be found here:

Error with address of parenthesized member function

Salpinx

Thanks for the link! ;) – Nf

@GLorieul Note that &std::cout isn't a function pointer at all, but the address of a global instance. – Devoid

The "explanation" is bogus. We are not forming a pointer to member here. – Sarraceniaceous

@πάνταῥεῖ yes indeed std::cout is an instance of class std::ostream accessible globally (i.e. declared as extern std::ostream std::cout). Yet the same problem as for member functions arise here (as I assume you suggested) – Nf

Answer 2

Before C++17 removed this restriction, the syntactic form of a template argument for a pointer or reference template parameter was restricted. N4140 [temp.arg.nontype]/1.3 says that it must be

expressed (ignoring parentheses) as & id-expression, where the id-expression is the name of an object or function, except that the & may be omitted if the name refers to a function or array and shall be omitted if the corresponding template-parameter is a reference

(std::cout) isn't an id-expression. It's a primary-expression.

The "(ignoring parentheses)" part was added by Core issue 773, and is apparently meant to permit (&i), not &(i).

Sarraceniaceous

This is a bit too technical for me… What is a "syntatic form"? – Nf