How to break const

[Timon Gehr]

On 06/18/2012 05:13 PM, deadalnix wrote:
> 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).

The whole problem is that the hidden parameter has changed. It is immutable!