Why doesn't C++ variadic template accept iostream value as a parameter? - c++

Question

I refer this link to implement variadic template:
#include <iostream>
#include <sstream>
// base case
void doPrint(std::ostream &out) {}
template <typename T, typename... Args>
void doPrint(std::ostream &out, T t, Args... args)
{
out << t; // add comma here, see below
doPrint(out, args...);
}
int main() {
// See how it works even better than varargs?
doPrint(std::cout, "Hola", " mundo ");
return 0;
}
But if I change function parameter from reference
void doPrint(std::ostream &out)
......
template <typename T, typename... Args>
void doPrint(std::ostream &out, T t, Args... args)
to value:
void doPrint(std::ostream out)
......
template <typename T, typename... Args>
void doPrint(std::ostream out, T t, Args... args)
I get the following build errors:
test.cpp: In function 'int main()':
test.cpp:16:40: error: 'std::basic_ostream<_CharT, _Traits>::basic_ostream(const std::basic_ostream<_CharT, _Traits>&) [with _CharT = char; _Traits = std::char_traits<char>]' is protected within this context
doPrint(std::cout, "Hola", " mundo ");
^
In file included from /usr/include/c++/7.1.1/iostream:39:0,
from test.cpp:1:
/usr/include/c++/7.1.1/ostream:391:7: note: declared protected here
basic_ostream(const basic_ostream&) = delete;
^~~~~~~~~~~~~
test.cpp:16:40: error: use of deleted function 'std::basic_ostream<_CharT, _Traits>::basic_ostream(const std::basic_ostream<_CharT, _Traits>&) [with _CharT = char; _Traits = std::char_traits<char>]'
doPrint(std::cout, "Hola", " mundo ");
^
In file included from /usr/include/c++/7.1.1/iostream:39:0,
from test.cpp:1:
/usr/include/c++/7.1.1/ostream:391:7: note: declared here
basic_ostream(const basic_ostream&) = delete;
^~~~~~~~~~~~~
test.cpp:8:6: note: initializing argument 1 of 'void doPrint(std::ostream, T, Args ...) [with T = const char*; Args = {const char*}; std::ostream = std::basic_ostream<char>]'
void doPrint(std::ostream out, T t, Args... args)
^~~~~~~
test.cpp: In instantiation of 'void doPrint(std::ostream, T, Args ...) [with T = const char*; Args = {const char*}; std::ostream = std::basic_ostream<char>]':
test.cpp:16:40: required from here
test.cpp:11:12: error: 'std::basic_ostream<_CharT, _Traits>::basic_ostream(const std::basic_ostream<_CharT, _Traits>&) [with _CharT = char; _Traits = std::char_traits<char>]' is protected within this context
doPrint(out, args...);
~~~~~~~^~~~~~~~~~~~~~
In file included from /usr/include/c++/7.1.1/iostream:39:0,
from test.cpp:1:
/usr/include/c++/7.1.1/ostream:391:7: note: declared protected here
basic_ostream(const basic_ostream&) = delete;
^~~~~~~~~~~~~
test.cpp:11:12: error: use of deleted function 'std::basic_ostream<_CharT, _Traits>::basic_ostream(const std::basic_ostream<_CharT, _Traits>&) [with _CharT = char; _Traits = std::char_traits<char>]'
doPrint(out, args...);
~~~~~~~^~~~~~~~~~~~~~
In file included from /usr/include/c++/7.1.1/iostream:39:0,
from test.cpp:1:
/usr/include/c++/7.1.1/ostream:391:7: note: declared here
basic_ostream(const basic_ostream&) = delete;
^~~~~~~~~~~~~
test.cpp:8:6: note: initializing argument 1 of 'void doPrint(std::ostream, T, Args ...) [with T = const char*; Args = {}; std::ostream = std::basic_ostream<char>]'
void doPrint(std::ostream out, T t, Args... args)
^~~~~~~
Per my understanding, use value may not have a good performance as reference, but it shouldn't cause compilation error. Why doesn't C++ variadic template accept iostream value as a parameter?

Answer 1

It's simply, because copy constructor in std::basic_ostream is deleted (by standard)
read about this here at cppreference.
So std::cout you can move or take by reference
example:
class A {
public:
A(const A& a) = delete;
};
void foo1(const A& a)
{
std::cout << "HEY1" << std::endl;
}
void foo2(const A a)
{
std::cout << "HEY2" << std::endl;
}
int main() {
A a{};
foo1(a);
// foo2(a);
return 0;
}
So in example you see that first function work fine when copy constructor is deleted, but if you uncomment foo2(a), that you get some information from compile e.g. (VS)
'A::A(const A &)': attempting to reference a deleted function

Answer 2

The explanation is in the first couple of lines of error messages.
std::ostream is actually a typedef for std::basic_ostream<char, std::char_traits<char> >.
Passing a std::ostream by value calls the copy constructor of the object, and the std::basic_ostream template has explicitly deleted the copy constructor so it cannot be called.