I have a method in a baseclass that needs the type passed to it for some type-related operations (lookup, size, and some method invocation). Currently it looks like this:
class base
{
template<typename T>
void BindType( T * t ); // do something with the type
};
class derived : public base
{
void foo() { do_some_work BindType( this ); }
};
class derivedOther : public base
{
void bar() { do_different_work... BindType( this ); }
};
However, I wonder if there's a way to get the caller's type without having to pass this so that my callpoint code becomes:
class derived : public base
{
void foo() { BindType(); }
};
Without the explicit this pointer. I know that I could supply the template parameters explicitly as BindType<derived>()
, but is there a way to somehow extract the type of the caller in some other way?
It will not work as you expect
The result of foo()
might be different of what you expect:
class derived : public base // <= YOU ARE IN CLASS DERIVED
{
public:
void foo() { BindType( this ); } // <= SO this IS OF TYPE POINTER TO DERIVED
};
The template paramter is deducted at compile time, so that it will be derived*
. If you would call foo()
from a class derived_once_more
derived from derived
, it would still use the type derived*
.
But you can get rid of the dummy parameter*
You may use decltype(this)
to represent the typename of a variable. It's still defined at compile time:
class base
{
public:
template<typename T>
void BindType( )
{
cout << typeid(T*).name()<<endl; // just to show the type
}
virtual ~base() {}; // for typeid() to work
};
class derived : public base
{
public:
void foo() { BindType<decltype(this)>( ); }
};
Edit: other alternatives
As template parameters need to be provided at compile-time and not a run time, you can use:
decltype (see above)