On Thursday, 14 March 2013 at 03:45:39 UTC, kenji hara wrote:
2013/3/14 deadalnix <[email protected]>
It is a tempting idea, but do not work. Consider :
void delegate() immutable a;
const void delegate() b = a;
This would be forbidden as covariance of function parameters
and
transitivity act in opposite directions.
You are misunderstanding about delegate type syntax and
transitivity.
1. This qualifier for delegate type should be added after
parameter types.
const void delegate() b; // bad, typeof(b) == const(void
deelgate())
void delegate() const b; // good, b can hold const method
After or before don't change anything here as type qualifier are
transitives. It means that a const delegate must have a const
context. Or, in code :
static assert(is(const delegate() == const delegate() const)); //
Pass
2. A delegate which has immutable context cannot implicitly
convertible to
const context delegate.
In general, if a delegate can implicitly convertible to
another, there
should be _contravariance_ of parameter types.
void delegate() immutable a;
void delegate() const b;
b = a; // bad
a = b; // good
As showed above, this is equivalent to my sample code in the
previous post, and it does break transitivity.
void delegate() const b; // Context of b is const (ie mutable or
immutable).
void delegate() immutable a; // Context of a is immutable
a = b; // a now contains a reference to possibly mutable data
considered as immutable by the type system.
This is *VERY* wrong.
