Copy Constructor Needed with temp object

Question

The following code only works when the copy constructor is available. When I add print statements (via `std::cout`) and make the copy constructor available it is not used (I assume there is so compiler trick happening to remove the unnecessary copy). But in both the output `operator <<` and the function `plop()` below (where I create a temporary object) I don't see the need for the copy constructor. Can somebody explain why the language needs it when I am passing everything by const reference (or what I am doing wrong). ``` #include <iostream> class N { public: N(int) {} private: N(N const&); }; std::ostream& operator<<(std::ostream& str,N const& data) { return str << "N
"; } void plop(std::ostream& str,N const& data) { str << "N
"; } int main() { std::cout << N(1); // Needs copy constructor (line 25) plop(std::cout,N(1)); // Needs copy constructor N a(5); std::cout << a; plop(std::cout,a); } ``` Compiler: [Alpha:~/X] myork% g++ -v Using built-in specs. Target: i686-apple-darwin10 C…

Accepted Answer

From http://gcc.gnu.org/gcc-3.4/changes.html

When binding an rvalue of class type to a reference, the copy constructor of the class must be accessible. For instance, consider the following code:

```
class A 
{
public:
  A();

private:
  A(const A&);   // private copy ctor
};

A makeA(void);
void foo(const A&);

void bar(void)
{
  foo(A());       // error, copy ctor is not accessible
  foo(makeA());   // error, copy ctor is not accessible

A a1;
  foo(a1);        // OK, a1 is a lvalue
}
```

This might be surprising at first sight, especially since most popular compilers do not correctly implement this rule (further details).

This will be fixed in C++1x by Core Issue 391.

Copy Constructor Needed with temp object

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