How to break const

[deadalnix]

Le 18/06/2012 16:55, Mehrdad a écrit :
> On Monday, 18 June 2012 at 14:48:37 UTC, deadalnix wrote:
>> Le 18/06/2012 16:44, Mehrdad a écrit :
>>> Interesting, making the delegate `pure' doesn't change anything either.
>>>
>>> So 'pure' doesn't let you "infer something just by looking at the code
>>> either", right?
>>
>> It does ! It tell you that the function have no side effect, and that
>> the function called with identical arguments will return identical
>> results.
>>
>> pure will not ensure constness or immutability. const and immutable
>> are made for that.
>>
>> The fact that D decouple purity and immutability is a very nice design
>> decision and is explained nicely here :
>> http://klickverbot.at/blog/2012/05/purity-in-d/
>
>
>
> Identical calls giving identical results? What?
>
>
> import std.stdio;
> struct S
> {
> this(int a)
> {
> this.a = a;
> this.increment = { return this.a++; };
> }
> int a;
> int delegate() pure increment;
> auto oops() const { return this.increment(); }
> }
> void main()
> {
> auto c = immutable(S)(0);
> writeln(c.oops()); // 0
> writeln(c.oops()); // 1
> writeln(c.oops()); // 2
> writeln(c.oops()); // 3
> writeln(c.oops()); // 4
> writeln(c.oops()); // 5
> }

They are not call with the same parameters. The hidden parameter have 
changed (I know this is tricky).