How to break const

[deadalnix]

Le 18/06/2012 17:16, Mehrdad a écrit :
> On Monday, 18 June 2012 at 15:13:33 UTC, deadalnix wrote:
>>> 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).
>
>
> Explain it however you want.
>
> The bottom line I'm getting at is, you can't re-order the calls, EVEN IF
> by "looking at them" you can tell they're pure @safe nothrow const...

Yes, but this isn't a problem with pure here, this is a problem with const.

The sample code should not compile.