Lazy evaluation

[Dicken]

Hi i have found this, and think it's dow to  Lazy /eager  evaluation; so  single  empty  list  = {} 

= let alist = {}  ,
tr = List.Transform( alist,(x)=> x {0} ), 
p_indi = alist {0} 
in tr

tr = empty , p_indi = error

But with a nested list ;

= let alist = {{}} ,
tr = List.Transform( alist,(x)=> x {0} ), 
p_indi = alist {0} 
in tr

tr = error, p_indi = empty 

 so if anyone would like to elaborate on why/ how this works, links to info   etc.  please reply,   I have no practical use for this
Just  for the brain. 

Richard. 

[Dicken]

so on   LIst.Transform(   {}  , (x)=>    ,     as the  list is empt theree is not list for  the function to work on, 
so  it  does nothing,  stays in that original  state,  where as  {} {0}  iis executed regardless, so = error ,   is this  along the right lines.

[krishnakanth240]

Yes @Dicken 

That's how I understand it as well. With List.Transform({}, (x) => x{0}) input list is empty, so there are no elements to pass into function. Since the function is never invoked, nothing attempts to evaluate x{0} and result remains an empty list. {}{0} is a direct indexing operation on an empty list, so it is evaluated immediately and returns an index out of range error

[AlienSx]

hi, (x) => x{0} is applied to each element of the list, not the list itself. That's why List.Transform with empty list { } returns the same empty list (nothing to transform) while List.Transform with { { } } returns ... no, not error but the list with single element {error}. And It has nothing to do with lazy / eager evaluation.

[krishnakanth240 · Accepted solution]

Hi @Dicken 

List.Transform operates on empty list {}, so transformation function (x) => x{0} is never executed resulting empty list.alist{0} directly attempts to access first element of an empty list which immediately raises an index out of range error

alist = {{}} contains one element (an empty list). List.Transform execute once passing {} as x and x{0} fails because inner list is empty, so tr becomes an error. alist{0} returns first element of the outer list which is the empty list {}, so no error