A delegate problem, create delegation in loop
Timon Gehr
timon.gehr at gmx.ch
Fri Jul 6 00:06:55 PDT 2012
On 07/06/2012 05:14 AM, lijie wrote:
> On Thu, Jul 5, 2012 at 4:26 PM, Denis Shelomovskij
> <verylonglogin.reg at gmail.com <mailto:verylonglogin.reg at gmail.com>> wrote:
>
> Different situation is for such C# loop:
> ---
> for (int i = 0; i < funcs.Length; ++i)
> {
> int t = i;
> funcs[i] = new MyFunc(() => System.Console.WriteLine(t));
> }
> ---
> where "t" is local for scope. Here C# behaves correctly, but D
> doesn't. This D loop
> ---
> foreach(i; 0 .. 5) {
> int t = i;
> functions ~= { printf("%d\n", t); };
> }
> ---
> prints "4" five times. It's Issue 2043:
> http://d.puremagic.com/issues/__show_bug.cgi?id=2043
> <http://d.puremagic.com/issues/show_bug.cgi?id=2043>
>
> How to distinguish which variables will be copied to the closure context?
>
They are not copied, they are stored there.
> I think this is a scope rule, in the previous code, there are three
> variables:
> 1. function arguments
> 2. loop variables
> 3. local variables
> Seems only function parameters is copied. In C#, local variables is
> copied. There are other rules? And why is the loop variable not local?
>
> Thanks.
>
> Best regards,
>
> -- Li Jie
It is simple. Variable declarations introduce a new variable. Closures
that reference the same variable will see the same values.
----
foreach(i; 0..3) { functions~={writeln(i);}; }
is the same as
for(int i=0;i<3;i++) { functions~={writeln(i);}; }
is the same as
{int i=0;for(;i<3;i++) { functions~={writeln(i);}; }}
is the same as
{
int i=0;
{ functions~={writeln(i);}; }
i++;
{ functions~={writeln(i);}; }
i++;
{ functions~={writeln(i);}; }
i++;
}
----
foreach(i; 0..3){ int j=i; functions~={writeln(j);}; }
is the same as
for(int i=0;i<3;i++){ int j=i; functions~={writeln(j);}; }
is the same as
{int i=0;for(i<3;i++){ int j=i; functions~={writeln(j);}; }
is the same as
{
int i=0;
{ int j=i; functions~={writeln(j);}; }
i++;
{ int j=i; functions~={writeln(j);}; }
i++;
{ int j=i; functions~={writeln(j);}; }
i++;
}
----
I think it is quite intuitive.
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