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我在试验时遇到了一个奇怪的问题CopyMemory。我有一段从Bytecomb复制的代码，只有在我将其放入函数时才有效：

我把这个放在开头：

Declare PtrSafe Sub CopyMemory Lib "kernel32" Alias "RtlMoveMemory" (pDest As Any, pSource As Any, ByVal ByteLen As Long)

工作版本：

Sub StringTest()
    Dim str1 As String
    Dim pstr1 As LongPtr

    str1 = "PowerVB"
    Debug.Print "Memory  : 0x"; Mem_ReadHex(pstr1 - 4, LenB(str1) + 6)

End Sub

Public Function Mem_ReadHex(ByVal Ptr As LongPtr, ByVal Length As Long) As String
    Dim bBuffer() As Byte, strBytes() As String, i As Long, ub As Long, b As Byte
    ub = Length - 1
    ReDim bBuffer(ub)
    ReDim strBytes(ub)
    CopyMemory bBuffer(0), ByVal Ptr, Length
    For i = 0 To ub
        b = bBuffer(i)
        strBytes(i) = IIf(b < 16, "0", "") & Hex$(b)
    Next
    Mem_ReadHex = Join(strBytes, "")
End Function

这个程序完美地打印出字符串在内存中的整个布局（前4个字节表示长度，后面是字符串内容，然后是2个字节的null）：

Memory  : 0x0E00000050006F00770065007200560042000000

现在，如果我将函数放入子程序中，它会崩溃：

Sub StringTest()
    Dim str1 As String
    Dim str2 As String
    Dim pstr1 As LongPtr

    str1 = "PowerVB"
    CopyMemory pstr1, ByVal VarPtr(str1), 8

    Dim bBuffer() As Byte, strBytes() As String
    ub = LenB(str1) + 5
    ReDim bBuffer(ub)
    ReDim strBytes(ub)
    CopyMemory bBuffer(0), ByVal pstr1 - 4, LenB(str1) + 6 'extra 4 bytes for string length, and 2 bytes of null characters
    For i = 0 To ub
        b = bBuffer(i)
        strBytes(i) = IIf(b < 16, "0", "") & Hex$(b) 'for 2 bytes, if the value < 16, then it only consists of one byte
    Next i
    Debug.Print Join(strBytes, "")
End Sub

我不明白。两个版本有什么区别？

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1 回答 1

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好的，我找到了解决方法：

CopyMemory bBuffer(0), ByVal pstr1 - 4, ub + 1

因为 CopyMemory 必须将 along作为第三个参数，所以我不能使用LenB(str1) + 6它，因为它可能是一个整数。

于 2018-05-29T04:00:48.323 回答