Re: Storing std::type_info
* Erik Wikstr?m:
On 2007-08-11 16:02, Daniel Kraft wrote:
Hi all,
I tried to store a std::type_info value in my code so I didn't have to
call typeid more than once when checking if a polymorphic object is of
one of certain types, like this:
#include <typeinfo>
class A
{
public:
virtual int foo() { return 42; }
};
class B : public A
{};
class C : public A
{};
void foo(A& obj)
{
std::type_info type(typeid(obj));
if(type==typeid(B)) { ... }
if(type==typeid(C)) { ... }
}
I found, however, that the copy-constructor of type_info was private;
is it save to store the value by reference:
const std::type_info& type(typeid(obj)); ?
This compiles fine, but I'm not sure whether this stores the reference
to a temporary object which is no longer valid when the comparison
happens.
Is is ok to do this with references? Or is there any other way to
store the type_info or do I have to insert typeid(obj) everywhere in
the ifs?
Yes, a const reference can be bound to a temporary.
Well, technically that doesn't happen here. The reference is bound to
the reference returned by typeid. Otherwise with the current standard
an accessible copy constructor would be required.
Or you can use & to get the address of the type_info object (typeid
returns a reference) but in that case you should use type_info's ==
operator and not compare addresses of type_info objects, i.e.
void foo(A& obj)
{
const std::type_info* type = &typeid(obj);
if(*type==typeid(B)) { std::cout << "B\n"; }
if(*type==typeid(C)) { std::cout << "C\n"; }
}
I think better to set the compiler more free to do its job without any
deep analysis, i.e., using the reference, or just typeid directly.
Which might actually be more efficient. However, this kind of thing
falls squarely into the Evil Premature Optimization category...
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