On Friday, 17 February 2012 at 02:49:40 UTC, Walter Bright wrote:
Given:
class A { void foo() { } }
class B : A { override pure void foo() { } }
This works great, because B.foo is covariant with A.foo,
meaning it can "tighten", or place more restrictions, on foo.
But:
class A { pure void foo() { } }
class B : A { override void foo() { } }
fails, because B.foo tries to loosen the requirements, and so
is not covariant.
Where this gets annoying is when the qualifiers on the base
class function have to be repeated on all its overrides. I ran
headlong into this when experimenting with making the member
functions of class Object pure.
So it occurred to me that an overriding function could
*inherit* the qualifiers from the overridden function. The
qualifiers of the overriding function would be the "tightest"
of its explicit qualifiers and its overridden function
qualifiers. It turns out that most functions are naturally
pure, so this greatly eases things and eliminates annoying
typing.
I want do to this for @safe, pure, nothrow, and even const.
I think it is semantically sound, as well. The overriding
function body will be semantically checked against this
tightest set of qualifiers.
What do you think?
I'm still not convinced about this apply to const. Consider this
example:
Initial code:
class A{
void foo(int) const;
void foo(float) const;
}
class B{
alias A.foo foo;
override void foo(int);
}
Revision to class A:
class A{
void foo(int);
void foo(int) const;
void foo(float);
void foo(float) const;
}
When the user recompiles, there will be no errors or warnings.
All uses of foo(int) through a const B will revert to using the
base class's implementation.
