Incomplete value at Symbol.cc:130

[Hans-Peter Sorge]

    Hi,
    
    I could not resist to experiment.....
    
          )clear
    CLEAR WS     
            A←0
            A.A←1
    DOMAIN ERROR+
            A.A←1
             ^ ^
    
          )clear
    CLEAR WS
          A.A←1
          A←0      ⍝ Just replacing content of A.    
         
          B.B←2
          A←B.B    ⍝ Generates a trace. - probably should be A.B ←→ 2   
    
    Incomplete value at Symbol.cc:130
      Addr:    0x1ce1990
      Rank:    0
      ..............
    
    
    ⍝ A structured variable of any depth can be replaced by an
    assignment to the root variable.
          A.B.C.D←4
          A
    .B:
    .B.C:
    .B.C.D: 4
    
          A←1   ⍝ Consistent with current assignment: replace variable
    content with RV. No DOMAIN ERROR.
          A
    1
    
    
    ⍝ Assignment working one time only ..
    
          A.B.C ←1
          X.Y.Z ←2
          A.B.C←X.Y
          A
    .B:
    .B.C:
    .B.C.Z: 2
    
    ⍝ should be repeatable:
          A.B.C←X.Y
    DOMAIN ERROR+
          A.B.C←X.Y
             ^   ^
    
    Proposal:
    
    Allow a non-leaf node to be replaced by a RV. it would then be
    consistent with how the current assignment works.    
    
    The assignment above "A.B.C ← X.Y" is actually an append ->
    A.B.C.Z rather than a replace,  what a "←" normally does.
    
    I think the assignment should result in A.B.C ← X.Y.  ←→ A.B.Z
    
    The append could be covered by the idiom  A.B.C. ← X.Y.   ←→ A.B.C.Z
                  
    .. just my thoughts.
      
          A.B.C ←1
            X.Y.Z ←2
            K.L.M ←3
      
            A.B.C ← X.Y K.L
      DOMAIN ERROR+
            A.B.C←X.Y K.L
               ^   ^
      'Idea:'
            A.B.C ← X.Y. K.L.   ←→ A.B.Z  A.B.M
      and:      
            A.B.C. ← X.Y. K.L.   ←→ A.B.C.Z  A.B.C.M 
            
    
    Basically saying:
    Using a trailing dots lets operators work with a structured name
    element.
    A.B. ←→ C
    
    given
        A.B.a←1 ⋄ A.B.b←2 ⋄ A.B.c←3
      X.Y.x←10 ⋄ X.Y.y←20 ⋄
            X.Y.z←30
          
        A.B. could result in structured name elements a b c. That is
        listing the leaf-node names.  
        A.B. ← X.Y.  could
            result in structured names A.B.x, A.B.y, A.B.z
      
    
    
    Without trailing dots we get the content of a variable.
    A.B.C  ←→ 1 
     
          A.B.a←1
          A.B.b←2
          A.B.c←3
          A
          A.B
    .a: 1
    
    .b: 2
    
    .c: 3
    
          A.B + 1
    DOMAIN ERROR
          A.B+1
           ^ ^
    ⍝ could be
          A.B + 1
    2 3 4
    
    
    Sorry for the lengthy note.  
    
    Best Regards
    Hans-Peter