Question

in the following code, does array of array A = B?

let A =  Array.init  3 (fun _ -> Array.init 2 (fun _ -> 0))
let defaultCreate n defaultValue = Array.init n (fun _ -> defaultValue)
let B = defaultCreate 3 (defaultCreate 2 0)

if I assign values to A and B, they are different ,what happened? thanks.

 for i = 0 to 2 do
     for j = 0 to 1 do
            A.[i].[j] <-i + j
            B.[i].[j] <-i + j
 printfn "%A vs %A" A  B

A = [|[|0; 1|]; [|1; 2|]; [|2; 3|]|] and B = [|[|2; 3|]; [|2; 3|]; [|2; 3|]|]

Answer 1

They are not the same.

Arrays are reference types, and are stored on the heap. When you create an array with another array as the default value, you are storing references to the same array, over and over again.

Numbers are another thing. They are immutable, and are stored by value, on the stack. So you can’t change the value of 1 to anything other than 1.

To create an “jagged” array, you need to call Array.init from inside the initializer to the first Array.init call, to create new arrays for each slot.

Also; You could use Array.create if you do want to have the same value in every slot. Be careful about reference types though.

let A = Array.init 3 (fun _ -> Array.create 2 0)