Android Question QUIZ - you can win... nothing ^_^

[LucaMs]

B4A, of course

Dim D As Double = 1250.60
D = D * 100
Log("D = " & D)

What the result will be?


And this?:

Dim D As Double = 1250.60
Dim L As Long = D * 100
Log("D = " & D)
Log("L =: " & L)

https://www.b4x.com/android/forum/threads/trasformare-un-importo-in-un-intero.58999/

Spoiler: VB.Net

  Private Sub Form1_Load(sender As Object, e As System.EventArgs) Handles Me.Load
      Dim D As Double = 1250.6
      D = D * 100
      Debug.Write(D)
      Application.Exit()
End Sub

Result: 125060

 

[Informatix]

B4A, of course

Dim D As Double = 1250.60
D = D * 100
Log("D = " & D)

What the result will be?


And this?:

Dim D As Double = 1250.60
Dim L As Long = D * 100
Log("D = " & D)
Log("L =: " & L)

https://www.b4x.com/android/forum/threads/trasformare-un-importo-in-un-intero.58999/

Spoiler: VB.Net

  Private Sub Form1_Load(sender As Object, e As System.EventArgs) Handles Me.Load
      Dim D As Double = 1250.6
      D = D * 100
      Debug.Write(D)
      Application.Exit()
End Sub

Result: 125060

It is due to Java, not to B4A.
Note that you will get the right result with a float.
As D * 100 is rounded down when converted to a long, the wrong result is pretty logical for L.
If you want to control the precision, never use floats or doubles, only the BigDecimal class (in Java) or as I saw in real applications using currencies, convert everything to long for your computations and use floats or doubles only for the display (by dividing the integer value).

 

[LucaMs]

It is due to Java, not to B4A.

I added that line after, because below I posted the same code but in VB.Net (someone could not distinguish them: thanks @Erel ).

Note that you will get the right result with a float.

I did not know this and "instinctively" I would have thought (at least) the vice-versa.

As D * 100 is rounded down when converted to a long, the wrong result is pretty logical for L.

Right... pretty .
Note that using VB.Net, the resultS are "pretty right" anyway:

    Private Sub Form1_Load(sender As Object, e As System.EventArgs) Handles Me.Load
        Dim D As Double = 1250.6
        Dim L As Long = D * 100
        D = D * 100
        Debug.WriteLine(D)
        Debug.WriteLine(L)
        Application.Exit()
    End Sub

125060
125060

If you want to control the precision, never use floats or doubles, only the BigDecimal class (in Java) or as I saw in real applications using currencies, convert everything to long for your computations and use floats or doubles only for the display (by dividing the integer value).

This is the goal of the member who asked the question (Italian forum).



Thank you, @Informatix