Re: How do I make the members of a parent class visible without using the "this" pointer?
peteymills@hotmail.com wrote:
It used to be that all the members (except those declared private)
of a parent class were visible to its descendants as if they had
been declared within the descendant class itself. Now some
brain-surgeon has changed the standard so that you have to add the
"this" pointer, requiring a load of extra, redundant typing for C++
code that uses inheritance.
This is only so if you have a templated base class. The compiler
cannot know that Base<T> contains the same members for all Ts. Someone
could add different specializations for some Ts.
This is fine if you are starting from
scratch, but what if you have legacy code that needs to be compiled
and you don't feel like picking through the code to add the "this"
pointer to every single inherited member? Is there any way I can
tell g++ to use the old rules (without reverting to an older
version)?
Not that I know of. Fixing the code seems like the better option. The
compiler will tell you (most of) the places anyway.
Bo Persson
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