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I have got a large header file (~10000 lines) which is auto-generated by a script/program out of my control.

In order to avoid to include this file in the declaration of my class, I forward declare the few types I need:

--myclass.h

namespace bl { class TypeA; class TypeB; } // Other stuff and myclass definition...

Now it turns out that TypeA and TypeB are not class-names, but are instead defined inside the auto-generated file as:

typedef SomeUnspecifiedClassName TypeA; typedef AnotherUnspecifiedClassName TypeB;

Where by SomeUnspecifiedClassName I mean that I can not forward-declare this type-name because it may change under various circumstances.

How can I forward-declare a typedef? (Can't use c++11)

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  • Can you use the class via references and pointers? stackoverflow.com/questions/553682/… Commented Apr 29, 2013 at 20:45
  • The answer to your specific question in your last sentence is already answered in the first related question: stackoverflow.com/questions/804894/… Commented Apr 29, 2013 at 20:46
  • Depending on where TypeA and TypeB are being used, you could consider refactoring myclass into a template. Commented Apr 29, 2013 at 20:53

3 Answers 3

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Simply - you can't. If you post your specific situations, though, there might be some workarounds to what you want to do.

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I worked around with some macros (a lot of, actually), by typedef-ing directly the "unspecified" types.
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You can write a script that extracts the ...UnspecifedClassName from the typedef lines in your auto-generated source file. Then, this script would be the basis of your own auto-generated header file that would forward declare those classes, as well as your typedef statements to them. Your myclass.h file can then #include that header file.

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1

One relatively decent solution that I have found useful on occasion is to create a trivial wrapper class:

Place in a header file:

class ClassA; // now use pointers and references to ClassA at will

Place in a source file:

#include <NastyThirdPartyHeader> class ClassA: public TypeA { public: ClassA(TypeA const &x): TypeA(x) {} ClassA &operator=(TypeA const &x) { TypeA::operator=(x); return *this; } };

Depending on your use case, that may be all you need.

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