01-14-2005, 01:57 PM
I donno much about windows program and just want to study how to call functions from DLL files.
I find an example http://www.newlisp.org/code/VB6.zip and modify it to
But the the display message is
I find an example http://www.newlisp.org/code/VB6.zip and modify it to
Code:
Option Explicit
Declare Function dllNewlispEval CDECL Lib "newlisp" Alias "newlispEvalStr" (ByVal LExpr As String) As Long
Declare Function LoadLibraryA Lib "kernel32" (ByVal s As String) As Long
Declare Sub FreeLibrary Lib "kernel32" (ByVal h As Long)
Declare Function lstrLen Lib "kernel32" Alias "lstrlenA" (ByVal lpString As Long) As Long
Declare Function lstrCpy Lib "kernel32" Alias "lstrcpyA" (ByVal lpString1 As String, ByVal lpString2 As Long) As Long
Dim Shared NewlispHandle As Long
Sub LoadNewLISP()
Dim mylib As String
Dim mypath As String
Dim hinst As Long
Print "in load"
mylib = "newlisp.dll"
NewlispHandle = LoadLibraryA(mylib)
Print "NewlispHandle=",NewlispHandle
If NewlispHandle = 0 Then
Print "can not find newlisp.dll"
End
End If
Print
End Sub
Sub UnloadNewLISP()
Print "in unload"
FreeLibrary NewlispHandle
Print "NewlispHandle=",NewlispHandle
End Sub
Function newlispEval(LispExpression As String) As String
Dim Result As String
Dim ResHandle As Long
Dim ResultCode As Long
Print "in Eval"
Print "LispExpression$=",LispExpression+chr$(0)
Print "dllNewlispEval=",dllNewlispEval("1")
ResHandle = dllNewlispEval(LispExpression)
Print "ResHandle=",ResHandle
Result = Space$(lstrLen(ByVal ResHandle))
Print "Result=",Result
lstrCpy ByVal Result, ByVal ResHandle
newlispEval = Result
Print "Result=",Result
Print
End Function
Dim LispExpression$
LispExpression$="(+ 1 2)"
LoadNewLISP()
Print newlispEval$(LispExpression$)
UnloadNewLISP()Quote:in loadwhat is wrong with the source? Thank you.
NewlispHandle= 1809055744
in Eval
LispExpression$=(+ 1 2)
dllNewlispEval= 3458168ResHandle= 3458168
Result=
Result=
in unload
NewlispHandle= 1809055744