Square-bracketed identifiers break the resolver

[ThunderFrame]

I haven't tried parsing this, but it might be nasty...

Option Explicit

Sub test()

  Dim s As String
  Dim i As Integer
  Dim rng As Range
  Dim a As Variant
  Dim v As Variant

  'Cell address B2
  s = [B2].Address
  s = [$B2].Address
  s = [$B$2].Address
  s = [B$2].Address

  'multi-cell
  s = [B:B].Address
  s = [B2:B4].Address

  'Cell address that can't be old Excel
  s = [XFD1048576]

  'Function name that could be a cell address
  s = E1

  'Function name that can't be a cell address
  s = [EEEE1]

'Excel formulas...
  s = [=TODAY()]
  s = [=B1 + B2]
  s = [=B1 & B2]
  v = [=And(true,true)]


  'Union operators
  s = [B1,B2,B3].Address
  s = [B1, B2, B3].Address

  'Intersection operators
  s = [B:B 2:2 B:C].Address

  'Intersection and union operators
  s = [B:B 2:2 B:C, D1 D:D].Address

  'Named Range
  i = [MyNamedRange].Rows.Count

  'Get range as array
  a = [B2:D4]

  'Get range formulas as array
  a = [B2:B4].Formula

  With [MyNamedRange]
    Set rng = .Cells
  End With

  Set rng = [MyNamedRange]
  Set rng = Sheet1.[MyNamedRange]
  Set rng = [Sheet1].[MyNamedRange]
  Set rng = [Sheet1!MyNamedRange]

End Sub

Public Function E1() As String
  E1 = [F1]
End Function

Public Function EEEE1() As String
  EEEE1 = "Test"
End Function

[ThunderFrame]

And how about:

  'Excel vs VBA function signatures
  v = [Log(10)] ' Returns 1
  v = Log(10)   ' Returns 2.30258509299405

  v = Round(4.5, 0) ' Returns 4 (Bankers rounding)
  v = [Round(4.5, 0)] ' Returns 5

  'Excel function vs VBA operator/keyword
  v = Not [Not(True)]

[retailcoder]

Hmm anything in square brackets is just going to be parsed as an (undeclared) identifier. It shouldn't affect anything - I'll test it out to be sure though.

[retailcoder]

It would be nice if an inspection/quickfix could turn them into explicit WorksheetFunctions and Range calls, but this would be a quite massive effort, for little gain. As I said the parser processes them as undeclared identifiers, which complicates things because any undeclared identifier is already a problem (there's no Declaration for it, so there's no IdentifierReference either) - i.e. note taken, but I'm not sure how to do anything about this.

[retailcoder]

@ThunderFrame ok, the code parses, but completely wrecks the resolver and using refactor/rename on any of the locals in the test procedure, breaks the code:

Sub test()

  Dim renamedS As String
  Dim renamedI As Integer
  Dim renamed As Range
  Dim renamedA As Variant
  Dim renamedV As Variant

  'Cell address B2
  renamedS = [B2].Address
  s = [$B2].Address
  s = [$B$2].Address
  s = [B$2].Address

  'multi-cell
  s = [B:B].Address
  s = [B2:B4].Address

  'Cell address that can't be old Excel
  s = [XFD1048576]

  'Function name that could be a cell address
  s = E1

  'Function name that can't be a cell address
  s = [EEEE1]

'Excel formulas...
  s = [=TODAY()]
  s = [=B1 + B2]
  s = [=B1 & B2]
  v = [=And(true,true)]


  'Union operators
  s = [B1,B2,B3].Address
  s = [B1, B2, B3].Address

  'Intersection operators
  s = [B:B 2:2 B:C].Address

  'Intersection and union operators
  s = [B:B 2:2 B:C, D1 D:D].Address

  'Named Range
  i = [MyNamedRange].Rows.Count

  'Get range as array
  a = [B2:D4]

  'Get range formulas as array
  a = [B2:B4].Formula

  With [MyNamedRange]
    Set rng = .Cells
  End With

  Set rng = [MyNamedRange]
  Set rng = Sheet1.[MyNamedRange]
  Set rng = [Sheet1].[MyNamedRange]
  Set rng = [Sheet1!MyNamedRange]

End Sub

...which is very scary at this point - this needs further investigation. My bet is on an exception that gets thrown in the resolver, which then stops trying to process the module any further.

[ThunderFrame]

And some Enum, for testing

Option Explicit

Public Enum FOO
  [BAR1] = 1
  [_BAR2] = 2 'Is hidden
End Enum

Sub test()

  Dim x

  x = BAR1   '1
  x = [BAR1] '1
  x = [BAS1] ' = Range("BAS1").Value
  x = [_BAR2] '2

End Sub

[retailcoder]

hmm the parse tree is broken, this will require a grammar fix.

For example, this expression should be picked up under the letStmt rule:

s = [B2].Address

But the LetStmtContext starts at line 12, and ends.... at line 63. The square-bracketed identifier parser rule is way too permissive:

ambiguousIdentifier : 
    (IDENTIFIER | ambiguousKeyword)+
    | L_SQUARE_BRACKET (.+)+ R_SQUARE_BRACKET
;

[retailcoder]

Fixed, at the lexer level.

Option Explicit

Sub Renamed6()

  Dim renamed1 As String
  Dim renamed2 As Integer
  Dim renamed3 As Range
  Dim renamed4 As Variant
  Dim renamed5 As Variant

  'Cell address B2
  renamed1 = [B2].Address
  renamed1 = [$B2].Address
  renamed1 = [$B$2].Address
  renamed1 = [B$2].Address

  'multi-cell
  renamed1 = [B:B].Address
  renamed1 = [B2:B4].Address

  'Cell address that can't be old Excel
  renamed1 = [XFD1048576]

  'Function name that could be a cell address
  renamed1 = Renamed8

  'Function name that can't be a cell address
  renamed1 = [EEEE1]

'Excel formulas...
  renamed1 = [=TODAY()]
  renamed1 = [=B1 + B2]
  renamed1 = [=B1 & B2]
  renamed5 = [=And(true,true)]


  'Union operators
  renamed1 = [B1,B2,B3].Address
  renamed1 = [B1, B2, B3].Address

  'Intersection operators
  renamed1 = [B:B 2:2 B:C].Address

  'Intersection and union operators
  renamed1 = [B:B 2:2 B:C, D1 D:D].Address

  'Named Range
  renamed2 = [MyNamedRange].Rows.Count

  'Get range as array
  renamed4 = [B2:D4]

  'Get range formulas as array
  renamed4 = [B2:B4].Formula

  With [MyNamedRange]
    Set renamed3 = .Cells
  End With

  Set renamed3 = [MyNamedRange]
  Set renamed3 = Renamed7.[MyNamedRange]
  Set renamed3 = [Sheet1].[MyNamedRange]
  Set renamed3 = [Sheet1!MyNamedRange]

End Sub

Public Function Renamed8() As String
  Renamed8 = [F1]
End Function

Public Function Renamed9() As String
  Renamed9 = "Test"
End Function

So, one problem remains: [Sheet1] should have been renamed to [Renamed7], right? Would it be correct to assume [Foo] is always equivalent to Foo whenever Foo resolves to a known identifier?

[retailcoder]

Ok so I've tweaked the resolver to treat [Foo] as Foo, and it works. It's also how VBA seems to handle it. I've entered the word TEST in cell A1, then proceeded to the immediate pane:

?[A1]
TEST

As expected, [A1] outputs TEST. Then I added a standard code module, and declared a Public A1 As String.

?[A1]

The immediate pane output was blank: the global variable takes precedence over the worksheet address in VBA's own identifier resolution algorithm, so treating [Foo] as Foo seems completely correct.

@ThunderFrame This code isn't valid however:

Set rng = [Sheet1!MyNamedRange]

The expression [Sheet1!MyNamedRange] returns Error 2023, so I'm not going to wreck my brains over that one.