Taking the address of a temporary object

Question

§5.3.1 Unary operators, Section 3

The result of the unary & operator is a pointer to its operand. The operand shall be an lvalue or a qualified-id.

What exactly does "shall be" mean in this context? Does it mean it's an error to take the address of a temporary? I was just wondering, because g++ only gives me a warning, whereas comeau refuses to compile the following program:

#include <string>

int main()
{
    &std::string("test");
}

g++ warning: taking address of temporary

comeau error: expression must be an lvalue or a function designator

Does anyone have a Microsoft compiler or other compilers and can test this program, please?

Answer 1

The word "shall" in the standard language means a strict requirement. So, yes, your code is ill-formed (it is an error) because it attempts to apply address-of operator to a non-lvalue.

However, the problem here is not an attempt of taking address of a temporary. The problem is, again, taking address of a non-lvalue. Temporary object can be lvalue or non-lvalue depending on the expression that produces that temporary or provides access to that temporary. In your case you have std::string("test") - a functional style cast to a non-reference type, which by definition produces a non-lvalue. Hence the error.

If you wished to take address of a temporary object, you could have worked around the restriction by doing this, for example

const std::string &r = std::string("test");
&r; // this expression produces address of a temporary

whith the resultant pointer remaining valid as long as the temporary exists. There are other ways to legally obtain address of a temporary object. It is just that your specific method happens to be illegal.