🚀 go-pugleaf

In article <1991Mar12.220532.897@objy.com> peter@objy.com writes:
>I think you are making this sound harder than it really is.

I think you are making this sound easier than it really is.
The three examples which follow build up to a C++ puzzle
which hopefully illustrates some of the nonsense which
C++ 2.1 may inflict on a lexical analyzer.

Example 1:
	struct V { int T; };
	struct A : virtual V { int T; };
	struct B : virtual V, A {
		B() { T = 1; }
	};

	Here A::T dominates V::T and thus "T = 1" is "this->A::T = 1;"

Example 2:
	struct V { struct T { }; };
	struct A : virtual V { T t; };

	Here V's T is in A's scope and thus A has a member "t" of type ::V::T.

Example 3:
	struct V { struct T { }; };
	struct A : virtual V { struct T { int ohno; }; };
	struct B : virtual V, A { T t; };

	Here A::T dominates V::T and thus "T t;" declares "::A::T t;"

	(Or does it: ARM 10.1.1, p. 205, reads, "A name B::f dominates
	a name A::f if its class B has A as a base." which would apply to
	type names.  However, if you read
		"When virtual bases are used, a single function, object,
		*TYPE*, or enumerator may be reached...this is not an
		ambiguity"
	and later
		"When virtual base classes are used, more than one
		function, object, or enumerator...dominates..."
	you might conclude that ARM leaves open whether or not dominance
	applies to type names.)

Puzzle:
	struct X { };
	struct V { struct T { }; };
	struct A : virtual V { int T; };
	struct B : virtual V, A { T(X); };

	??? What does "T(X);" mean ???

My answer follows the ^L.

What implications do you suppose this has on the structure of a lexical
analyzer?

Comments?
Jan Gray

My answer follows...

If A::T did not dominate V::T, then "T(X);" would declare X to be a ::V::T.
Since A::T dominates V::T, T is not a type, therefore "T(X);" means
"int T(X);"