0

according to the open standard (http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2006/n2027.html)

This code should give an error:

A&  a_ref3 = A();  // Error!
A&& a_ref4 = A();  // Ok

However, I am able to do the first line just fine. And the second line actually errors for me. I am compiling with Visual C++14 in Visual Studio.

class foo {

};

int main()
{

    foo x;
    foo() = x; // Valid

    foo& fr = foo(); // Valid

    string&& foo = "hey"; // Valid r reference

    foo&& frr = foo(); // Error!

    return 0;
}

So I guess my question is that is foo() an lvalue? And if so, why does Stroustrup say that you cannot assign a reference type to it?

David Parker
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    The first line should be an error if you disable language extensions. I'm not sure what's up with the second one not working. – chris Oct 05 '15 at 00:41
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    `foo&& frr = foo();` fails because you've redeclared `foo` as a `std::string&&`, change the name of the variable in the line above. – user657267 Oct 05 '15 at 00:44
  • Can you fine gents please copy your message to the answer section so I can accept it? It looks like language extensions are not complying to ISO C++. – David Parker Oct 05 '15 at 00:47

1 Answers1

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foo& fr = foo(); works in VC++ because the compiler allows it, clang and gcc both reject this syntax.

foo&& frr = foo(); on the other hand is an error because you've redeclared foo and are trying to call operator() on a std::string, which isn't defined.

Community
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user657267
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    VC++ will warn you as well, if you set the warning level high enough. It's a level 4 warning for historical reasons. – Bo Persson Oct 05 '15 at 07:05