On Friday, 10 October 2014 at 02:38:42 UTC, Walter Bright wrote:
For functions, const-as-storage-class applies to the function
symbol. And if it is misused, the compiler will very likely
complain about a mismatched type.
Breaking this adds a special case inconsistency, besides
breaking existing code.
Come on that is the same bogus reason every time. const here do
not apply to the function but to its hidden, implicit parameter.
And that is actually a problem.
const void delegate() dg; // dg should be const, and there is no
way to qualify the implicit parameter.
You obviously can't pretend you don't know this as:
const int foo() {}
Gives you an error because you have no this pointer to make
const. So let's not pretend that this const actually qualify the
function when everybody knows it doesn't.
