Question

我正在尝试通过ML32中的直接逐位传输将vb 2015 Integer更改为vb 2015 UInteger。从vb2015开始，我在dll中调用vc ++ 2015函数，该函数使用内联汇编程序来实现更改。

通过这样做，我可以采用负整数，如-633593090 =＆amp; HDA3C22FE = 11011010001111000010001011111110b

并将其更改为UInteger = 11011010001111000010001011111110b =＆amp; HDA3C22FE = 3661374206

与看似简单的“uTest1 = CUInt（iTest1）”相反，它会为iTest1负面抛出System.OverflowException。

我的代码运行正常，但我对可能会有一点改进感兴趣。

在vb 2015中，我声明了对vc ++ 2015 dll的访问：

' vb Integer --> c++ signed long --> c++ unsigned long --> vb UInteger -- Works!
'   by direct copy of bits through x86 ML32
'   4 bytes = 32 bits
<DllImport("StringTest.dll", EntryPoint:="bitConvert", SetLastError:=True, CharSet:=CharSet.Ansi, ExactSpelling:=True, CallingConvention:=CallingConvention.Cdecl)>
Private Shared Function bitConvertTester(ByRef varSInt As Integer, ByRef varUInt As UInteger) As Integer
    ' Must be "Shared"
    ' Must be "ByRef"
    ' Do not try varSInt As "Signed Integer" or varUInt As "Unsigned Integer" --> Syntax Error
    ' Leave the body of the function empty
End Function

我使用这个vb 2015代码实际调用dll中的vc ++ 2015函数：

    ' Third Method - Partial - Works!
    Dim pCTest As Color
    Dim iTest As Integer
    Dim uTest As UInteger
    Dim returnCode As Integer
    pCTest = Color.FromArgb(&HDA, &H3C, &H22, &HFE)
    iTest = pCTest.ToArgb
    System.Diagnostics.Debug.Write(iTest.ToString + vbCrLf)
    System.Diagnostics.Debug.Write(Hex(iTest) + vbCrLf)

    ' vb Integer --> c++ signed long --> c++ unsigned long --> vb UInteger
    '   by direct copy of bits through x86 ML32
    '   4 bytes = 32 bits
    ' Note: Function prototype is "ByRef", but this MUST 
    '   NOT go in this test call (causes a syntax error).
    returnCode = bitConvertTester(iTest, uTest)

    System.Diagnostics.Debug.Write(uTest.ToString + vbCrLf)
    System.Diagnostics.Debug.Write(Hex(uTest) + vbCrLf)

这是vc ++ 2015 dll函数本身：

    #define EXPORT_VB extern "C" __declspec(dllexport)
    // vb Integer --> c++ signed long --> c++ unsigned long --> vb UInteger -- Works!
    //   by direct copy of bits through x86 ML32
    //   4 bytes = 32 bits
    EXPORT_VB long __cdecl bitConvert(signed long *varSInt, unsigned long *varUInt)
    {
        // Intermediate variables are required for ML32 access.
        signed long varSInt1;
        unsigned long varUInt1;
        varSInt1 = *varSInt;
        __asm
        {
            mov EAX, varSInt1
            mov varUInt1, EAX
        }
        *varUInt = varUInt1;
        return 1;                   // Success Code
    }

我使用变量ByRef从vb 2015调用vc ++ 2015 dll函数，以便dll函数可以使用指针通过* varSInt和* varUInt根据需要更改vb 2015变量。

我的问题是：在vc ++ __asm块中，有没有办法让ML32代码直接访问* varSInt和* varUInt，而无需通过varSInt1和varUInt1中间变量？

BTW，vb 2015代码使用：

 Imports System
 Imports System.IO
 Imports System.Text
 Imports System.Runtime.InteropServices

vc ++ 2015代码使用：

 #include "stdafx.h"
 #include <stdexcept>
 #include <string.h>

（我发现Microsoft文档令人沮丧的一件事是未能包含使用文档描述的功能所需的“Imports”或“#include”语句列表。

Answer 1

我意识到你说你只是把它作为一个练习，但有一种相对简单的方法可以在VB（或C＃）中获得你想要的结果。

Function ToUInt(value As Integer) As UInteger
    Return CUInt(If(value < 0, UInteger.MaxValue + value + 1, value))
End Function

vb整数到c ++到ML32到c ++到vb UInteger

1 个答案: