Слишком мало аргументов lisp ошибка

---

@Shneuph wrote:

Kent — I still don’t fully understand the differences between C: functions and ones w/o the «C:» and how they are used differently.  I do know that some built-in autolisp functions have «optional» arguements which you cannot have with a C: function.  To get around this I have something like C:Symbol (P1 P2 P3 Poptional1 Poptional2 / etc etc)

Then in the lisp I just test to see if Poptional1 or Poptional2 are nil and if so adjust appropritely.  Then call the function like

(C:Symbol «p1» «p2» «p3» «po1» nil)

In the lisp it will have:

(if Poptional1

  (do this)
);if

(if Poptional2

  (do this)

);if

---

Things defined with the C:, e.g. (defun C:Whatever… are commands [I assume that’s what the C stands for], though they are also sometimes referred to as routines, and their command names can be typed at the Command: prompt line without parentheses, as with other commands, or put in menu items, scripts, etc.  But since they’re not native AutoCAD commands, they can’t be used in a Lisp (command) function if they need to be called from inside other Lisp routines, but need to be in parentheses with the C: included.  They’re not supposed to have arguments, though that doesn’t seem to be absolute [see Message 4].  But they can have localized variables.

Things defined without the C:, e.g. (defun Whatever… are usually called routines, or sub-routines, or maybe some other things.  [Sometimes someone will refer to one as a function, but that has the potential to be confused with what are properly called functions: those listed in the AutoLISP Reference, including (defun) itself.]  You can’t type their bare names at the Command: prompt line, though you can type them there included in parentheses, but usually they are used from within other routines, similarly to the way AutoLISP functions are used.  They can have arguments [though they don’t always need to], as well as localized variables.

In your example, if you’re going to use it by calling it from inside another Lisp function/routine, I would just define it as a routine/sub-routine, not as a command, since the C: isn’t doing anything for you:(defun Symbol (P1 P2 P3 Poptional1 Poptional2 / etc etc)

and use it the way functions/routines/sub-routines are used:

(Symbol p1 p2 p3 po1 nil)

[Note no quotation marks around the arguments — I’m assuming those are variables holding some kind of value, and not actually text strings.]

Kent Cooper, AIA

Please, help me with this program to toggle multiple wire numbers.

Error: too few arguments…

Private Sub ToggleWireNum()

Dim x As AcadEntity

Dim blk As AcadBlock

On Error Resume Next

For Each x In ThisDrawing.ModelSpace

    If TypeName(x) = «IAcadLine» Then

        ThisDrawing.SetVariable «USERS1», x.Handle

        WDCommand («(setq return (c:ace_get_wnum (handent (getvar «»USERS1″»))))» & vbCr)

        WDCommand («(setvar «»USERS2″» (car return))» & vbCr)

        If Len(ThisDrawing.GetVariable(«USERS2»)) > 0 Then

            WDCommand («(setq en (cadr return))» & vbCr)

            WDCommand («(c:ace_toggle_inline en)» & vbCr)

        End If

    End If

Next x

End Sub

Public Sub WDCommand(Command As String)

‘ This function calls lisp function

ThisDrawing.SendCommand (Command)

End Sub

Public Sub WDForceStart()

Call WDCommand(«(if(not wd_load)(if(setq x(findfile «»wd_load.lsp»»))(load x)))(wd_load)» & vbCr)

End Sub

Кто бы ни писал этот код, похоже, не знал о progn оператор.

То есть, если мы хотим оценить несколько выражений e1, e2,… из-за побочного эффекта, который они производят, нам не нужно этого делать:

;; wrap expressions in a dummy lambda and apply empty arg list to it
(apply (lambda () e1 e2 ...) nil)

Мы можем просто написать это:

(progn e1 e2 ...)

Это по-прежнему оставляет нам странный запах кода, который выглядит так:

(setq str (progn .... (setq str value)))

Код присваивает переменную strдважды с одинаковым значением. Глубоко вложенный(setq str value) ставит value в strа затем возвращает это значение в качестве результата. Это последнее выражениеprogn (первоначально из lambda), и это значение также возвращается. Тогда внешнийsetq расточительно хранит это в strеще раз. Нам просто нужно одно или другое:

;; set str as the side effect of the last form in the
;; progn; also yield that value.
(progn e1 e2 ...(setq str value))

;; assign value yielded from progn to str, then also
;; yield that value.
(setq str (progn e1 e2 ... value))