Android Question How can i find the value of Gravity

Scantech · Mar 5, 2016

B4X:

'    Gravity.NO_GRAVITY                = 0
'    Gravity.CENTER_HORIZONTAL        = 1
'    Gravity.LEFT                    = 3
'    Gravity.RIGHT                    = 5
'    Gravity.CENTER_VERTICAL            = 16
'    Gravity.CENTER                    = 17
'    Gravity.TOP                        = 48
'    Gravity.BOTTOM                    = 80
'    Gravity.FILL                    = 119

    Dim x As Int
   
    x = Gravity.BOTTOM + Gravity.LEFT

'how can i find x that contains Gravity.Left?

Cableguy · Mar 5, 2016

I dont understand the question!
what do you want to know?
X? or Gravity.Left?

Gravity.left = x - Gravity.botom

sorex · Mar 5, 2016

if you want to check if your view has gravity.left set you can use

B4X:

If Bit.And(l.Gravity,3)=3 then

or maybe

B4X:

If Bit.And(l.Gravity,gravity.left)=gravity.left then

Scantech · Mar 5, 2016

sorex said:
if you want to check if your view has gravity.left set you can use

B4X:

If Bit.And(l.Gravity,3)=3 then

or maybe

B4X:

If Bit.And(l.Gravity,gravity.left)=gravity.left then

What is l?

You probably meant this

B4X:

If Bit.And(x, 3) = 3 Then

This will not work because the Gravity values is not bitmapped in 1,2,4,8,16,32,64,128.

Example why it will not work.

B4X:

    Dim x As Int
'   
    x = Gravity.BOTTOM + Gravity.FILL
    If Bit.And(x, 3) = 3 Then
        Log("YES")
    End If

It returns Yes.

Scantech · Mar 5, 2016

Cableguy said:
I dont understand the question!
what do you want to know?
X? or Gravity.Left?

Gravity.left = x - Gravity.botom

That is not going to work. The x can have any of the Gravity combination and I want to detect Gravity.left(3) is being used.

I don't think there is a mathematical equation for this problem. I need to find other ways. Value 17 can represent (Gravity.Center) or (Gravity.Center_Vertical + Gravity.Center_Horizontal) and that example is why it will be impossible to detect which one is being used.

Scantech · Mar 5, 2016

I just realized some of those gravity properties can not be mixed together. So If I want to detect Gravity.Left I can perhaps use this.

B4X:

    Select Case x
        case 3, 19, 51, 83: log("YES") 
    End Select

3 = Gravity.Left
19 = Gravity.Left + Gravity.Center_Vertical
51 = Gravity.Left + Gravity.Top
83 = Gravity.Left + Gravity.Bottom

Those are the only combinations allowed?

Cableguy · Mar 5, 2016

I would say yes, since some gravity values affect both vertical as horizontal behaviours, like gravity.center

I would say that any combined value is unique and therefore no 2 different combinations can give the same value

mc73 · Mar 6, 2016

Supposing that the one who sets gravity does it "correctly", you can have

B4X:

 Dim gravityLeftSelected As Boolean=(x mod 16=3)

Scantech · Mar 6, 2016

mc73 said:
Supposing that the one who sets gravity does it "correctly", you can have

B4X:

Dim gravityLeftSelected As Boolean=(x mod 16=3)

Nice! 1 line of code

Android Question How can i find the value of Gravity

Scantech

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