在VBA中正确声明DWord [英] Correct declaration of DWord in VBA

问题描述

我正在使用API​​函数，该函数返回DWORD

I'm using an API function which returns a DWORD

因为我想在LoWord和HiWord上使用智能，而不是使用Long:

Because I want intellisense on the LoWord and HiWord, rather than using a Long:

Declare Sub myAPI(ByRef outVariable As Long)

...如WinAPI的此列表-> VBA数据类型转换中所建议的，我正在使用一种类型:

...as suggested in this list of WinAPI -> VBA datatype conversions, I'm using a type:

Public Type DWORD 'same size as Long, but intellisense on members is nice
    '@Ignore IntegerDataType
    LoWord As Integer
    '@Ignore IntegerDataType
    HiWord As Integer
End Type

Declare Sub myAPI(ByRef outVariable As DWORD)

然而，RubberDuck的 IntegerDataType 检查提醒我在32位系统上

However RubberDuck's IntegerDataType inspection reminded me that on 32 bit systems VBA converts 2-byte Integers to 4-byte Longs internally, so I'm wondering whether my DWORD declaration is really 4 consecutive bytes as expected, or 8.

我对指针和位&不够熟悉.字节在脑海中描绘出了什么情况，但是我想API某种程度上只知道填充每个部分的下半部分，因为我一直在从API中获得我期望的结果(我认为).

I'm not familiar enough with pointers and bits & bytes to picture in my head what's going on, but I imagine the API somehow knows to fill only the lower half of each part, as I've been getting the results I expect (I think) from the API.

推荐答案

您定义的类型为4个字节，因为Integer为2个字节.

Your user defined type is 4 bytes is size, because Integer is 2 bytes in size.

您可以自己检查:

Dim dw as DWORD
Dim size as Integer
size = LenB(dw)

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