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?
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)
.
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
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