How to replace all items in an array of doubles with the same value

Question

So lets say we have an array of doubles like this one that is later used for other stuff.

double[] myArray = new double[25];

How would I go about replacing all the values in that array with a set value?

`var array = Enumerable.Range(0, 25).Select(_ => 0d).ToArray();` — Damith, Dec 02 '16 at 09:24

Bathsheba · Accepted Answer · 2016-12-02T09:19:52.143

6

There are flashier ways, but

for (int i = 0; i < myArray.Length; ++i){
    myArray[i] = foo; /*the new value*/
}

is clear and simple.

edited Dec 02 '16 at 09:19

answered Dec 02 '16 at 09:18

Bathsheba

wouldn't it better if we define it like `i < myArray.Length`? – Ali Bahrami Dec 02 '16 at 09:19
1

@AliBahraminezhad: Of course you're correct. I've pinched that. – Bathsheba Dec 02 '16 at 09:20

Manfred Radlwimmer · Answer 2 · 2016-12-02T09:21:44.713

6

The "flashy" way (simply create a new array):

var array = Enumerable.Repeat(value, count).ToArray();

Or

Array.ConvertAll(array, e => value);

edited Dec 02 '16 at 09:21

answered Dec 02 '16 at 09:19

Showoff ;-) Have an upvote. – Bathsheba Dec 02 '16 at 09:20
I think instead of `count` better to write myArray.Length? – Ali Bahrami Dec 02 '16 at 09:21
1

`Repeat` is less efficient than the for-loop approach and `ConvertAll` is made for a different purpose(convert the type), so a bit misleading. It's also using the for-loop[[source](https://referencesource.microsoft.com/#mscorlib/system/array.cs,f4af030bcf8c8991,references)]. – Tim Schmelter Dec 02 '16 at 09:24
Flashy ways will create new instance of array instead of updating existed one – Fabio Dec 02 '16 at 09:25
@TimSchmelter Totally aware of that ^^. I just wanted to complete Bathsheba's answer since he mentioned *flashy* ways. There are of course countless other ways of abusing built in methods for this purpose. – Manfred Radlwimmer Dec 02 '16 at 09:30

2 Answers2