Re: Why .dup not work with multidimensional arrays?

2015-05-08 Thread Ali Çehreli via Digitalmars-d-learn

On 05/07/2015 07:39 PM, Dennis Ritchie wrote:

On Friday, 8 May 2015 at 02:23:23 UTC, E.S. Quinn wrote:

It's because arrays are references types, and .dup is a strictly
shallow copy, so you're getting two outer arrays that reference
the same set of inner arrays. You'll have to duplicated each of
the inner arrays yourself if you need to make a deep copy.


Thank you. It really works :)

-
import std.stdio;

void main() {

 auto c = [[[1, 2, 3], [4, 5, 6, 7, 8]],
   [[9, 10], [11, 12, 13]]];

 auto d = [[c[0][0].dup, c[0][1].dup],
   [c[1][0].dup, c[1][1].dup]];

 d[0][1][1 .. $ - 1] *= 3;

 writeln(c = , c);
 // [[[1, 2, 3], [4, 5, 6, 7, 8]],
 //  [[9, 10], [11, 12, 13]]] // OK
 writeln(d = , d);
 // [[[1, 2, 3], [4, 15, 18, 21, 8]],
 //  [[9, 10], [11, 12, 13]]] // OK
}
-
http://ideone.com/kJVUhd

Maybe there is a way to create .globalDup for multidimensional arrays?


In D, everything is possible and very easy. :p I called it deepDup:

import std.stdio;
import std.traits;
import std.range;
import std.algorithm;

auto deepDup(A)(A arr)
if (isArray!A)
{
static if (isArray!(ElementType!A)) {
return arr.map!(a = a.deepDup).array;

} else {
return arr.dup;
}
}

void main()
{
auto c = [[[1, 2, 3], [4, 5, 6, 7, 8]],
  [[9, 10], [11, 12, 13]]];

auto d = c.deepDup;

d[0][1][1 .. $ - 1] *= 3;

writeln(c = , c);
// [[[1, 2, 3], [4, 5, 6, 7, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
writeln(d = , d);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
}

Ali



Re: Why .dup not work with multidimensional arrays?

2015-05-08 Thread Dennis Ritchie via Digitalmars-d-learn

On Friday, 8 May 2015 at 06:30:46 UTC, Ali Çehreli wrote:
In D, everything is possible and very easy. :p I called it 
deepDup:


import std.stdio;
import std.traits;
import std.range;
import std.algorithm;

auto deepDup(A)(A arr)
if (isArray!A)
{
static if (isArray!(ElementType!A)) {
return arr.map!(a = a.deepDup).array;

} else {
return arr.dup;
}
}

void main()
{
auto c = [[[1, 2, 3], [4, 5, 6, 7, 8]],
  [[9, 10], [11, 12, 13]]];

auto d = c.deepDup;

d[0][1][1 .. $ - 1] *= 3;

writeln(c = , c);
// [[[1, 2, 3], [4, 5, 6, 7, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
writeln(d = , d);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
}

Ali


Thank you. In D it's really easy :)
Recursion, which works with the lambda map looks fine.


I was a little question: why static int idx variable declared 
within a function deepDup takes the values 1, 1, 1, 2, 2, 3, 4, 
as opposed to a global variable static int idx, which receives 
the expected value of 1, 2, 3, 4, 5, 6, 7 ?


-
import std.stdio,
   std.range,
   std.traits,
   std.algorithm;

// static int idx; // 1, 2, 3, 4, 5, 6, 7 // OK

auto deepDup(A)(A arr)
if (isArray!A)
{
static int idx; // 1, 1, 1, 2, 2, 3, 4 // Why is this happening?
++idx;
writeln(visited);
static if (isArray!(ElementType!A)) {
writeln(ifIdx = , idx);
writeln(ifArr = , arr);
return arr.map!(a = a.deepDup).array;

} else {
writeln(elseIdx = , idx);
writeln(elseArr = , arr);
return arr.dup;
}
}

void main() {

auto a = [[[1, 2, 3], [4, 5, 6, 7, 8]],
  [[9, 10], [11, 12, 13]]];

auto b = a.deepDup;

b[0][1][1 .. $ - 1] *= 3;

writeln(\nResualt: );

writeln(a = , a);
// [[[1, 2, 3], [4, 5, 6, 7, 8]],
//  [[9, 10], [11, 12, 13]]]
writeln(b = , b);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]]
}
-
http://ideone.com/mAHZyO


Re: Why .dup not work with multidimensional arrays?

2015-05-08 Thread Dennis Ritchie via Digitalmars-d-learn

On Friday, 8 May 2015 at 15:13:14 UTC, Ali Çehreli wrote:

On 05/08/2015 08:05 AM, Dennis Ritchie wrote:

 why static int idx variable declared within a
 function deepDup takes the values 1, 1, 1, 2, 2, 3, 4, as
opposed to a
 global variable static int idx, which receives the expected
value of 1,
 2, 3, 4, 5, 6, 7 ?

That's because every template instance is a different type (or 
implementation). Just like the static variables of foo and bar 
are separate below, so are the static variables of t!int and 
t!float:


void foo()
{
static int i;
}

void bar()
{
static int i;
}

void t(T)()
{
static int i;
}

Ali


Thankы. Now everything is clear.


Re: Why .dup not work with multidimensional arrays?

2015-05-08 Thread Ali Çehreli via Digitalmars-d-learn

On 05/08/2015 08:05 AM, Dennis Ritchie wrote:

 why static int idx variable declared within a
 function deepDup takes the values 1, 1, 1, 2, 2, 3, 4, as opposed to a
 global variable static int idx, which receives the expected value of 1,
 2, 3, 4, 5, 6, 7 ?

That's because every template instance is a different type (or 
implementation). Just like the static variables of foo and bar are 
separate below, so are the static variables of t!int and t!float:


void foo()
{
static int i;
}

void bar()
{
static int i;
}

void t(T)()
{
static int i;
}

Ali



Re: Why .dup not work with multidimensional arrays?

2015-05-08 Thread Chris via Digitalmars-d-learn

On Friday, 8 May 2015 at 06:30:46 UTC, Ali Çehreli wrote:

On 05/07/2015 07:39 PM, Dennis Ritchie wrote:

On Friday, 8 May 2015 at 02:23:23 UTC, E.S. Quinn wrote:
It's because arrays are references types, and .dup is a 
strictly
shallow copy, so you're getting two outer arrays that 
reference
the same set of inner arrays. You'll have to duplicated each 
of

the inner arrays yourself if you need to make a deep copy.


Thank you. It really works :)

-
import std.stdio;

void main() {

auto c = [[[1, 2, 3], [4, 5, 6, 7, 8]],
  [[9, 10], [11, 12, 13]]];

auto d = [[c[0][0].dup, c[0][1].dup],
  [c[1][0].dup, c[1][1].dup]];

d[0][1][1 .. $ - 1] *= 3;

writeln(c = , c);
// [[[1, 2, 3], [4, 5, 6, 7, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
writeln(d = , d);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
}
-
http://ideone.com/kJVUhd

Maybe there is a way to create .globalDup for multidimensional 
arrays?


In D, everything is possible and very easy. :p I called it 
deepDup:


import std.stdio;
import std.traits;
import std.range;
import std.algorithm;

auto deepDup(A)(A arr)
if (isArray!A)
{
static if (isArray!(ElementType!A)) {
return arr.map!(a = a.deepDup).array;

} else {
return arr.dup;
}
}

void main()
{
auto c = [[[1, 2, 3], [4, 5, 6, 7, 8]],
  [[9, 10], [11, 12, 13]]];

auto d = c.deepDup;

d[0][1][1 .. $ - 1] *= 3;

writeln(c = , c);
// [[[1, 2, 3], [4, 5, 6, 7, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
writeln(d = , d);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
}

Ali


Nice one. I have the same problem in one of my modules. I might 
use the above code henceforth.


Re: Why .dup not work with multidimensional arrays?

2015-05-07 Thread Dennis Ritchie via Digitalmars-d-learn

On Friday, 8 May 2015 at 02:23:23 UTC, E.S. Quinn wrote:

It's because arrays are references types, and .dup is a strictly
shallow copy, so you're getting two outer arrays that reference
the same set of inner arrays. You'll have to duplicated each of
the inner arrays yourself if you need to make a deep copy.


Thank you. It really works :)

-
import std.stdio;

void main() {

auto c = [[[1, 2, 3], [4, 5, 6, 7, 8]],
  [[9, 10], [11, 12, 13]]];

auto d = [[c[0][0].dup, c[0][1].dup],
  [c[1][0].dup, c[1][1].dup]];

d[0][1][1 .. $ - 1] *= 3;

writeln(c = , c);
// [[[1, 2, 3], [4, 5, 6, 7, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
writeln(d = , d);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
}
-
http://ideone.com/kJVUhd

Maybe there is a way to create .globalDup for multidimensional 
arrays?


Re: Why .dup not work with multidimensional arrays?

2015-05-07 Thread E.S. Quinn via Digitalmars-d-learn

It's because arrays are references types, and .dup is a strictly
shallow copy, so you're getting two outer arrays that reference
the same set of inner arrays. You'll have to duplicated each of
the inner arrays yourself if you need to make a deep copy.

On Friday, 8 May 2015 at 02:15:38 UTC, Dennis Ritchie wrote:

Hi,
Should the method .dup work with multidimensional arrays for 
copying?


-
import std.stdio;

void main() {

auto a = [1, 2, 3];
auto b = a.dup;

b[] *= 2;
writeln(a = , a); // [1, 2, 3] // OK
writeln(b = , b); // [2, 4, 6] // OK


auto c = [[[1, 2, 3], [4, 5, 6, 7, 8]],
  [[9, 10], [11, 12, 13]]];

auto d = c.dup;

writeln(d[0][1][1 .. $ - 1] = ,
 d[0][1][1 .. $ - 1]);

d[0][1][1 .. $ - 1] *= 3;

writeln(c = , c);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]] // wrong
writeln(d = , d);
// [[[1, 2, 3], [4, 15, 18, 21, 8]],
//  [[9, 10], [11, 12, 13]]] // OK
}
-
http://ideone.com/Ddtm47

I thought the slice of the array c[0][1][1 .. $ - 1] = [5, 6, 
7] not had to change to [15, 18, 21] by multiplying by 3.