Defer in D

[Nick Treleaven via Digitalmars-d]

On 20/03/2016 16:18, Bauss wrote:
>>> defer((int i) => writeln(i), i);
>>
>> If it was defer(() => writeln(i)), would that do the same? (Dpaste
>> won't seem to let me edit your example on my tablet).

(or on my laptop)

> No that's not the same, since the reference to i does not change when
> doing:
> defer(() => writeln(i))
>
> However with defer((int i) => writeln(i), i) you pass the value of i to
> the lambda expression which stores the value in a new reference.
>
> The problem with the first one is the value will always be the last
> value of i.

Thanks. So we need the lambda to capture args (below) rather than 
capturing i at the callsite:

     void opCall(ARGS...)(void delegate(ARGS) call, ARGS args)
     {
         stack.put(() => call(args));
     }

The closure above allocates its copy of args on the heap, instead of the 
callsite closure which captures i by reference.