Question

我正在尝试在VB6应用程序中编写对GetProcessImageFileName的调用，但是得到了......

Run-time error '453':
    Can't find DLL entry point GetProcessImageFileName in PSAPI.DLL

我理解它可以在文档here中的PSAPI.DLL中找到。

我的代码看起来像......

Public Declare Function GetProcessImageFileName Lib "PSAPI.DLL" _
        (ByVal hProcess As Long, _
         lpImageName As String, _
         ByVal nSize As Long) As Long

Public Sub MySub()
Dim name_length As Long
Dim image_name As String
...fill in process handle...
name_length = GetProcessImageFileName(process_handle, image_name, 1024)

有谁知道我应该在这做什么？我在Windows XP上运行。

编辑正如JosephH所建议，我已将代码更改为使用GetProcessImageFileNameA，因此......

Public Declare Function GetProcessImageFileNameA Lib "PSAPI.DLL" _
        (ByVal hProcess As Long, _
         lpImageName As String, _
         ByVal nSize As Long) As Long

和

name_length = GetProcessImageFileNameA(process_handle, image_name, 1024)

执行此操作（与W版本相同）会导致程序和VB6开发环境崩溃，因此在某处隐藏了另一个问题。

Answer 1

应该是GetProcessImageFileNameA或GetProcessImageFileNameW。接受字符串作为参数的大多数Windows API函数（除了GetProcAddress）都有两个原型，一个带有ANSI（带有A后缀），另一个带有unicode（带有W后缀）

Public Declare Function GetProcessImageFileName Lib "PSAPI.DLL" Alias "GetProcessImageFileNameA" _
    (hProcess As Long, _
     ByVal lpImageName As String, _
     nSize As Long) As Long

name_length = GetProcessImageFileNameA(process_handle, image_name, 1024)

我在哪里可以找到VB6中的GetProcessImageFileName？

1 个答案: