* Jason Kraftcheck:
> The same syntax is accepted by the only other C++ compiler I have
> access too: Sun's. I guess what I don't understand is why, if I
> create a temporary by explicitly calling the copy constructor, that
> temporary is treated as an rvalue.
The standard says so in section 3.10 ("Lvalues and rvalues"):
| 6 An expression which holds a temporary object resulting from a cast to
| a nonreference type is an rvalue (this includes the explicit creation
| of an object using functional notation (5.2.3)).
The type you're casting to is not a reference type, so it's an rvalue
according to section 5.4 ("Explicit type conversion (cast notation)").
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