"Jarrett Billingsley" <jarrett.billings...@gmail.com> wrote in message news:mailman.635.1233675301.22690.digitalmars-d-le...@puremagic.com... > On Tue, Feb 3, 2009 at 10:14 AM, Frank Benoit > <keinfarb...@googlemail.com> wrote: >> >> arr = arr[ 0 .. lowerBound ] ~ arr[ upperBound .. $ ]; >> > > That's simple enough, but inefficient. > > Something like this: > > import std.c.string; // or import tango.stdc.string; > > T[] erase(T)(ref T[] arr, size_t idx) > { > if(arr.length == 0) > throw new Exception("FAILCOPTER"); > > if(idx < arr.length - 1) > memmove(&arr[idx], &arr[idx + 1], T.sizeof * (arr.length - idx - > 1)); > > arr.length = arr.length - 1; > return arr; > }
Let's see if I understand memmove.. The way it's used here, it copies the tail of an array onto that same array, only starting one index earlier, thus removing the undesired element? Neat. However I just realized that order does not matter in the array I'm using, so I guess I could use something like: arr[ind] = arr[$-1]; arr.length = arr.length -1; Seeing how this only copies once, I'm guessing this is more efficient?
