🚀 go-pugleaf

matt@godzilla.EECS.Berkeley.EDU (Matt Austern) writes:

>fjh@munta.cs.mu.OZ.AU (Fergus Henderson) writes:
>
>> But no standard-conforming code defines `operator[] (int, int)'.
>> If Borland makes `a[ x, y ]' call `operator[] (int, int)' in
>> the case that `a' has an `operator[] (int, int)', it won't
>> affect any standard-conforming code.
>
>I don't think that's true.  a[x, y] is a legal C++ expression: it
>is equivalent to a[(x, y)], and, if evaluating x has no side 
>effects, it is equivalent to a[y].

You missed the point.  `a[x, y]' is a legal C++ expression, but
`operator[] (int, int)' is not a legal C++ declaration.  If Borland
were to change the meaning of `a[x, y]' only if the code declares
an `operator[]' with two arguments, then that cannot affect any
standard-conforming code, since no standard-conforming code can
possibly contain such a declaration.

	// prog1.cpp - strictly conforming C++ code
	struct Foo {
		int operator [] (int);
	}
	void bar(int x, int y) {
		Foo foo;
		foo[x,y];	// In any standard-conforming compiler, this
				// must be equivalent to `foo[y]'.
	}

	// prog2.cpp - this code is not strictly conforming C++
	struct Foo {
		int operator [] (int, int);
				// The above declaration is "ill-formed".
				// The compiler must issue a diagnostic.
	}
	void bar(int x, int y) {
		Foo foo;
		foo[x,y];	// Once the above diagnostic has been issued,
				// the compiler is free to do anything it likes
				// with this code, including calling
				// `Foo::operator[](int, int)'.
	}

-- 
Fergus Henderson              | Designing grand concepts is fun;
fjh@cs.mu.oz.au               | finding nitty little bugs is just work.
http://www.cs.mu.oz.au/~fjh   | -- Brooks, in "The Mythical Man-Month".
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