[Closed] More than one object equal portions of the problem

SelA=Selection as array
 NumA=0.0
 DistanceA=100
 For i = 1 to SelA.count do
 (
     SelA[i].pos.x = NumA + DistanceA+ (SelA[i].pos.x-SelA[i].min.x)
     NumA= SelA[i].Max.x
 )

Hi stefan Thank you.
Maybe I did not say understand,
The first object’s position and the last object’s position is fixed (Unchanged).

SelA=Selection as array
ObjXDim=0.0
For i = 2 to (SelA.count-1) do
(
	ObjXDim=ObjXDim+(SelA[i].max.x-SelA[i].min.x)
)
XStart=SelA[1].max.x
For i = 2 to (SelA.count-1) do
(
	FromMinToPivot=(SelA[i].pos.x-SelA[i].min.x)
	SelA[i].pos.x=XStart+(((SelA[SelA.count].min.x - SelA[1].max.x)-ObjXDim)/(SelA.count-1))+FromMinToPivot
	XStart=SelA[i].max.x
)

Hi stefan
After my test, your script is not correct.
Please see picture !

here is a basic idea:

fn x_spaceObjects nodes space:20 pos:[0,0,0] = 
(
	for node in nodes do
	(
		size = (node.max - node.min)
		node.center = [pos.x + size.x*0.5, pos.y, pos.z + size.z*0.5]
		pos.x += size.x + space
	)
)
/*
(
	delete objects
	classes = #(box, teapot, sphere, cylinder, cone)
	nodes = for k=1 to 20 collect 
	(
		node = classes[random 1 classes.count] scale:([1,1,1]*(random 0.5 2.0)) wirecolor:(random white black)
	)
	x_spaceObjects nodes
)
*/

anything else do yourself.

it is for multiple objects
select 3-100 objects run script and it
-takes x width of objects.
-subtract it from space between first and last obj selected. Divide it by number of objects
-moves objects one by one and leaves equal space between them
it depends on selection order,
but I probably do not understand so…