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I have this code (question from exam) and there are many errors in the code, here is the code:

 class A { int z; static int y; public: A() {(*this).incrementY();} int x () { y=y+2; return z;} static int getY(){ return z;} void incrementY() {y++} }; class B: A { public: B (): A() {(*this).incrementY();}; }; class C : public B { public: C(): B() {(*this).incrementY;() } }; int main() { A::x(); A a; a.getY(); A::getY(); B b; C c; C::getY(); b.x(); }

There is a private inheritance. This means that all the methods and variables in B become private to any class that inherits from B?

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  • This code is terrible, but it's best that you check it with any free C++ compiler. You'll get the whole story. Commented Mar 10, 2009 at 16:51

3 Answers 3

Reset to default
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There are so many that you cannot but get tired. Here are some:

void incrementY() {y++}

No ; after y++.

24. A::x();

Non static member cannot be invoked via a class name.

25. A a;

No definition of static member y. If y was a const this would've been okay. This is a bit tricky so I'll quote.

9.4.2 Static data members

  1. The declaration of a static data member in its class definition is not a definition and may be of an incomplete type other than cv-qualified void [...]

  2. If a static data member is of const effective literal type, its declaration in the class definition can specify a constant-initializer brace-or-equal-initializer with an initializer-clause that is an integral constant expression. A static data member of effective literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a constant-initializer brace-or-equal-initializer with an initializer-clause that is an integral constant expression. In both these cases, the member may appear in integral constant expressions. The member shall still be defined in a namespace scope if it is used in the program and the namespace scope definition shall not contain an initializer.

26. a.getY(); 27. A::getY();

Illegal reference to non-static member A::z.

Taken care of by first observation.

28. B b; 29. C c; 30. C::getY();

getY() is a private member of B, not visible to C, let alone be public.

31. b.x();

The member function x() inherited from A is private.

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2 Comments

I voted you up. But please note that the quote is from the working paper of the next Standard. So it mentions things not appearing in current C++ in effect (literal types, constexpr).
@litb: I usually mention that (n2798, C++0x draft). Thanks for pointing this out.
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You are correct there (although it actually means that all A's methods become inaccessible).

However, there are a few other problems: 

A::x() // will not work as x is not static. a.getY() // will not work as getY() is static. C::getY() // Cannot access getY()

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Yes, that is correct, although you could just compile it with any number of online C++ compilers to verify.

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