Fix sig fig check for large number when decimal is present

Fixes #30

Author: timhall

JsonConverter.bas

@@ -468,6 +468,7 @@ End Function
 Private Function json_ParseNumber(json_String As String, ByRef json_Index As Long) As Variant
     Dim json_Char As String
     Dim json_Value As String
+    Dim json_IsLargeNumber As Boolean
 
     json_SkipSpaces json_String, json_Index
 
@@ -483,8 +484,10 @@ Private Function json_ParseNumber(json_String As String, ByRef json_Index As Lon
             ' This can lead to issues when BIGINT's are used (e.g. for Ids or Credit Cards), as they will be invalid above 15 digits
             ' See: http://support.microsoft.com/kb/269370
             '
-            ' Fix: Parse -> String, Convert -> String longer than 15 characters containing only numbers and decimal points -> Number
-            If Not JsonOptions.UseDoubleForLargeNumbers And Len(json_Value) >= 16 Then
+            ' Fix: Parse -> String, Convert -> String longer than 15/16 characters containing only numbers and decimal points -> Number
+            ' (decimal doesn't factor into significant digit count, so if present check for 15 digits + decimal = 16)
+            json_IsLargeNumber = IIf(InStr(json_Value, "."), Len(json_Value) >= 17, Len(json_Value) >= 16)
+            If Not JsonOptions.UseDoubleForLargeNumbers And json_IsLargeNumber Then
                 json_ParseNumber = json_Value
             Else
                 ' VBA.Val does not use regional settings, so guard for comma is not needed

specs/Specs.bas

@@ -96,12 +96,14 @@ Public Function Specs() As SpecSuite
     End With
 
     With Specs.It("should handle very long numbers as strings (e.g. BIGINT)")
-        JsonString = "[123456789012345678901234567890, 1.123456789012345678901234567890]"
+        JsonString = "[123456789012345678901234567890, 1.123456789012345678901234567890, 123456789012345, 1.23456789012345]"
         Set JsonObject = JsonConverter.ParseJson(JsonString)
 
         .Expect(JsonObject).ToNotBeUndefined
         .Expect(JsonObject(1)).ToEqual "123456789012345678901234567890"
         .Expect(JsonObject(2)).ToEqual "1.123456789012345678901234567890"
+        .Expect(JsonObject(3)).ToEqual 123456789012345#
+        .Expect(JsonObject(4)).ToEqual 1.23456789012345
 
         JsonConverter.JsonOptions.UseDoubleForLargeNumbers = True
         JsonString = "[123456789012345678901234567890]"