Re: Why .dup not work with multidimensional arrays?
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?
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?
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?
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?
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?
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?
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.
