how to add dip with variables

[Aysic4Bandroid]

hi all,

This may be an embarrassingly simple thing but im puzzled as to how i go about using the dip measurement with a variable name so when i have an integer variable such as:

dim leftedge as int
leftedge=112

spnOpenfile.left = leftedge dip

Doing so highlights 'dip' as an error it works leaving it off but is there a way to explicitly add 'dip' to an int variable name?? :BangHead:

 

[NJDude]

Like this:

dim leftedge as int
leftedge=112dip

 

[Aysic4Bandroid]

ok,
should have tried it first, i did it but didnt run it cos it kind of looks so wrong like concatenating an int to a string

thanks for that, NJDude

 

[Erel]

You can also apply the dip unit to a variable directly:

Button1.Left = SomeVar * 1dip

 

[Mike Henry]

I don't know if anyone is still reading this old thread, but I'd like to add a comment.
It appears DIP returns an integer, so if DIP is 0.9 then 1DIP will return 0.
The only way I got DIP to work with a variable is to convert it to a double before hand.
Like this:

   Dim dipf As Double
   dipf = 100dip / 100
   V.Left = ReScale(N) * dipf

 

[klaus]

You could use:

V.Left = DipToCurrent(ReScale(N))

Best regards.

 

[Erel]

@Mike you are correct. The code I posted in post #5 is wrong. It should have been:

Button1.Left = SomeVar * 100dip / 100