如果不放入函数中，CopyMemory将崩溃

Question

在试验CopyMemory时，我遇到了一个奇怪的问题。我有一段从Bytecomb复制过来的代码，只有当我把它放入一个函数中时才能正常工作：

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内存中字符串的整个布局(前4个字节表示长度，然后是字符串内容，然后是2个字节的null)：

Memory  : 0x0E00000050006F00770065007200560042000000

现在，如果我把函数放到sub中，它就会崩溃：

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

我还是不明白。这两个版本有什么不同？

Answer 1

好的，我找到了解决方法：

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

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