Hi,
unlike the examples I found on D.learn regarding this topic I want to remove an element from an dynamic array without knowing the index.

A few questions :

//Remove element try #1
T[] remove(T,E)(T[] arr, E element)
{
  for(uint i = 0;i < arr.length-1;i++)
  {
    if (arr[i] == element)
      return i==arr.length-1 ? arr[0..$-1].dup : arr[0..i] ~ arr[i+1 .. $];		
  }
  // return T[]    ???????????????????
}

or is it preferable to use :
//Remove element try #2
void remove(T,E)(inout T[] arr, in E element)
{}

Given :
class myclass
{
  this(int x){}
}

myclass[] objarr;

auto a = new myclass(1);
objarr ~= a;
auto b = new myclass(2);
objarr ~= b;
auto c = new myclass(3);
objarr ~= c;

objarr.remove(b);

Do I need opEquals within myclass ?

Let's assume that the passed parameter is f.i. a priority index. In case  that I want to sort objarr ordered by this /priority index/, is toHash() the override of choice ?

Can you please help me to improve the code ?
Many thanks in advance, Bjoern

This may help clear your mind some:

import std.stdio: writefln;
import std.string: format;

void removeBang(T)(ref T[] arr, T el) {
    foreach(i, x; arr)
        if (x == el) {
            arr = arr[0 .. i] ~ arr[i+1 .. $];
            break;
        }
}

class C(bool withEqual) {
    int x;
    this(int x) {
        this.x = x;
    }
    string toString() {
        return format("C(%d)", x);
    }
    static if (withEqual) {
        bool opEquals(C other) {
            return this.x == other.x;
        }
    }
}

void main() {
    int[] a = [1, 2, 3, 4, 1];
    writefln(a);
    a.removeBang(4);
    writefln(a);
    writefln;

    alias C!(true) C1;
    auto b = new C1(2);
    C1[] oa = [new C1(1), b, new C1(3)];
    removeBang(oa, b);
    writefln(oa);
    removeBang(oa, new C1(1));
    writefln(oa);
    writefln;

    alias C!(false) C2;
    auto b2 = new C2(2);
    C2[] oa2 = [new C2(1), b2, new C2(3)];
    removeBang(oa2, b2);
    writefln(oa2);
    removeBang(oa2, new C2(1));
    writefln(oa2);
}


Output:

[1,2,3,4,1]
[1,2,3,1]

[C(1),C(3)]
[C(3)]

[C(1),C(3)]
[C(1),C(3)]

Bye,
bearophile

Thanks,
much more than I've expected from an answere!
and beside,thanks to your answere I know that I've to learn,learn,... :)
Bjoern