digitalmars.D - A delegate problem, create delegation in loop

lijie (15/39) Jul 04 2012 }

bearophile (11/23) Jul 05 2012 import std.stdio;

lijie (4/30) Jul 05 2012 Thanks bearophile.

Denis Shelomovskij (40/60) Jul 05 2012 This program behaves as expected. Just like C# program that will give

lijie (14/33) Jul 05 2012 How to distinguish which variables will be copied to the closure context...

Timon Gehr (38/70) Jul 06 2012 It is simple. Variable declarations introduce a new variable. Closures

lijie (2/38) Jul 06 2012 Understood, thanks.

lijie <cpunion gmail.com> writes:

Hi,

My test code:


 import std.stdio;
 void main() {
     void delegate()[] functions;
     foreach (i; 0 .. 5) {
         functions ~= {
             printf("%d\n", i);
         };

    }
     foreach (func; functions) {
         func();
     }
 }


output:

$ dmd

DMD64 D Compiler v2.059

...

$ ./main
 5
 5
 5
 5
 5


Seems like all delegations shared a stack variable, I think delegation must
copy the value into its context.

I can avoid it:

    void delegate() createDelegate(int i) {
         void exec() {
             printf("%d\n", i);
         }
         return &exec;
     }
     foreach (i; 0 .. 5) {
         functions ~= createDelegate(i);
     }


But hope for improve it.


Best regards,

-- Li Jie

Jul 04 2012

"bearophile" <bearophileHUGS lycos.com> writes:

lijie:

 import std.stdio;
 void main() {
     void delegate()[] functions;
     foreach (i; 0 .. 5) {
         functions ~= {
             printf("%d\n", i);
         };

     }
     foreach (func; functions) {
         func();
     }
 }



import std.stdio;

void main() {
     void delegate()[] functions;

     foreach (i; 0 .. 5)
         functions ~= ((int j) => { printf("%d\n", j); })(i);

     foreach (func; functions)
         func();
}

Bye,
bearophile

Jul 05 2012

lijie <cpunion gmail.com> writes:

On Thu, Jul 5, 2012 at 3:36 PM, bearophile <bearophileHUGS lycos.com> wrote:

 lijie:


  import std.stdio;
 void main() {
     void delegate()[] functions;
     foreach (i; 0 .. 5) {
         functions ~= {
             printf("%d\n", i);
         };

     }

     foreach (func; functions) {
         func();
     }
 }


 import std.stdio;

 void main() {
     void delegate()[] functions;

     foreach (i; 0 .. 5)
         functions ~= ((int j) => { printf("%d\n", j); })(i);

     foreach (func; functions)
         func();
 }

Thanks bearophile.

Best regards,

-- Li Jie

Jul 05 2012

Denis Shelomovskij <verylonglogin.reg gmail.com> writes:

05.07.2012 9:54, lijie пишет:
 Hi,

 My test code:

     import std.stdio;
     void main() {
          void delegate()[] functions;
          foreach (i; 0 .. 5) {
              functions ~= {
                  printf("%d\n", i);
              };

          }
          foreach (func; functions) {
              func();
          }
     }


 output:
     5
     5
     5
     5
     5


the same output:
---
delegate void MyFunc();

class Program
{
     static void Main()
     {
         var funcs = new MyFunc[5];

         for (int i = 0; i < funcs.Length; ++i)
             funcs[i] = new MyFunc(() => System.Console.WriteLine(i));

         foreach (var f in funcs)
             f();
     }
}
---

because "i" is the same for every iteration. Different situation is for 

---
for (int i = 0; i < funcs.Length; ++i)
{
     int t = i;
     funcs[i] = new MyFunc(() => System.Console.WriteLine(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

I'm not posting workaround here because bearophile already did it.

-- 
Денис В. Шеломовский
Denis V. Shelomovskij

Jul 05 2012

lijie <cpunion gmail.com> writes:

On Thu, Jul 5, 2012 at 4:26 PM, Denis Shelomovskij <
verylonglogin.reg gmail.com> wrote:



 ---
 for (int i = 0; i < funcs.Length; ++i)
 {
     int t = i;
     funcs[i] = new MyFunc(() => System.Console.WriteLine(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?

I think this is a scope rule, in the previous code, there are three
variables:
1. function arguments
2. loop variables
3. local variables

copied. There
are other rules? And why is the loop variable not local?

Thanks.

Best regards,

-- Li Jie

Jul 05 2012

Timon Gehr <timon.gehr gmx.ch> writes:

On 07/06/2012 05:14 AM, lijie wrote:
 On Thu, Jul 5, 2012 at 4:26 PM, Denis Shelomovskij
 <verylonglogin.reg gmail.com <mailto:verylonglogin.reg gmail.com>> wrote:


     ---
     for (int i = 0; i < funcs.Length; ++i)
     {
          int t = i;
          funcs[i] = new MyFunc(() => System.Console.WriteLine(t));
     }
     ---

     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

 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.

Jul 06 2012

lijie <cpunion gmail.com> writes:

On Fri, Jul 6, 2012 at 3:06 PM, Timon Gehr <timon.gehr gmx.ch> wrote:

 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.

Understood, thanks.

Jul 06 2012

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digitalmars.D - A delegate problem, create delegation in loop