lambda function with "capture by value"

[Simon Bürger via Digitalmars-d-learn]

On Saturday, 5 August 2017 at 18:22:38 UTC, Stefan Koch wrote:
> On Saturday, 5 August 2017 at 18:19:05 UTC, Stefan Koch wrote:
>> On Saturday, 5 August 2017 at 18:17:49 UTC, Simon Bürger wrote:
>>> If a lambda function uses a local variable, that variable is 
>>> captured using a hidden this-pointer. But this capturing is 
>>> always by reference. Example:
>>>
>>>     int i = 1;
>>>     auto dg = (){ writefln("%s", i); };
>>>     i = 2;
>>>     dg(); // prints '2'
>>>
>>> Is there a way to make the delegate "capture by value" so 
>>> that the call prints '1'?
>>>
>>> Note that in C++, both variants are available using
>>>   [&]() { printf("%d", i); }
>>> and
>>>    [=]() { printf("%d", i); }
>>> respectively.
>>
>> No currently there is not.
>
> and it'd be rather useless I guess.
> You want i to be whatever the context i is a the point where 
> you call the delegate.
> Not at the point where you define the delegate.

No, sometimes I want i to be the value it has at the time the 
delegate was defined. My actual usecase was more like this:

	void delegate()[3] dgs;
	for(int i = 0; i < 3; ++i)
		dgs[i] = (){writefln("%s", i); };


And I want three different delegates, not three times the same. I 
tried the following:

	void delegate()[3] dgs;
	for(int i = 0; i < 3; ++i)
	{
		int j = i;
		dgs[i] = (){writefln("%s", j); };
	}

I thought that 'j' should be considered a new variable each time 
around, but sadly it doesn't work.