Re: Arrays - Inserting and moving data

[MattCodr]

On Thursday, 9 February 2012 at 18:30:22 UTC, Ali Çehreli wrote:
On 02/09/2012 03:47 AM, MattCodr wrote:
I have a doubt about the best way to insert and move (not 
replace) some
data on an array.
For example,

In some cases if I want to do action above, I do a loop moving 
the data
until the point that I want and finally I insert the new data 
there.


In D I did this:

begin code
.
.
.
int[] arr = [0,1,2,3,4,5,6,7,8,9];

arr.insertInPlace(position, newValue);
arr.popBack();
.
.
.
end code


After the insertInPlace my array changed it's length to 11, so 
I use
arr.popBack(); to keep the array length = 10;

The code above is working well, I just want know if is there a 
better way?

Thanks,

Matheus.

Most straightforward that I know of is the following:

   arr = arr[0 .. position] ~ [ newValue ] ~ arr[position + 1 
.. $];

But if you don't actually want to modify the data, you can 
merely access the elements in-place by std.range.chain:

import std.stdio;
import std.range;

void main()
{
   int[] arr = [0,1,2,3,4,5,6,7,8,9];
   immutable position = arr.length / 2;
   immutable newValue = 42;

   auto r = chain(arr[0 .. position], [ newValue ], 
arr[position + 1 .. $]);
   writeln(r);
}

'r' above is a lazy range that just provides access to the 
three ranges given to it. 'arr' does not change in any way.

Ali

Hi Ali,

You gave me a tip with this "chain" feature.

I changed a few lines of your code, and it worked as I wanted:


import std.stdio;
import std.range;
import std.array;

void main()
{
    int[] arr = [0,1,2,3,4,5,6,7,8,9];
    immutable position = arr.length / 2;
    immutable newValue = 42;

    auto r = chain(arr[0 .. position], [ newValue ], arr[position 
.. $-1]);
    arr = array(r);

    foreach(int i; arr)
        writefln("%d", i);
}


Thanks,

Matheus.