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.
