How to call a C++ DLL from VB6

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Hello All,

I am trying to call a function in a C++ DLL from VB6.

The function call in the C++ DLL is aslfollow:

>DLLFUNC(int) MDNSa35dbd7v2(int service, int AC0_CryptoKeyTable, unsigned char* seed, unsigned char* key) {
return (a35dbd7v2(service, AC0_CryptoKeyTable, seed, key));
}

Where #define VBDLLFUNC(TypeName) DLLEXPORT TypeName __stdcall

What would the actual declare look like in VB6

According to the customer, if I send 5 bytes of 11 for the seed or 1111111111

I sould get back 5 bytes of 22 for the key or 2222222222

Please advise

Thank you

What I have tried:

I tried alot of Googling but have not found anything related to my case.

Posted 9-Dec-19 7:12am

gary hagerty

Updated 9-Dec-19 20:06pm

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Richard MacCutchan 9-Dec-19 14:49pm

You need to talk to the people who created the library.

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KarstenK · Accepted Answer · 2019-12-09T20:07:00

Solution 1

Read this great article Calling C++ DLLs from VC++ and VB for details.

Tip: create an own C++ dll with the same interface to test and verify your interface code.

Posted 9-Dec-19 20:07pm

KarstenK

Comments

Maciej Los 10-Dec-19 4:01am

5ed!

gary hagerty 10-Dec-19 9:43am

Hello Sir,

Thank you very much for your reply.

Please be aware that I am a C++ newbie.

I use the following declare when calling the function above:

Declare Function MDNSa35dbd7v2 Lib "C:\NewSP\VCSrc\MDTestDLL\Debug\MDTestDLL.dll" (ByVal service As Long, ByVal CryptoKeyTable As Long, ByVal Seed As String, ByVal Key As String) As Long

When I call the function, I do so as follows:

Dim lRetVal As Long
Dim sKey As String
Dim sSeed As String

sSeed = Chr(11) & Chr(11) & Chr(11) & Chr(11) & Chr(11)
lRetVal = MDNSa35dbd7v2(&H1, &H3C, sSeed, sKey)

I would like to accomplish the following in order to reach my goal:

1. How can I verify sSeed to make sure that it is equal to what I passed in?

2. In the C++ codes I set key = seed and then display the result in VB6 but I get an empty string, what am I doing wrong?

3. What is the VB6 equivalent of unsigned char*?

4. Does the * mean that the value is being passed in by reference?

Thank you

Gary