初始化 SUSI API dll 失败

发布于 2025-01-18 07:21:54 字数 1169 浏览 0 评论 0原文

我正在尝试使用提供给我的Susi API DLL。根据他们的文件，第一步是致电Susilibinitialize。当调用它时，当我期望获得SUSI_STATUS_Initialized作为结果时，我似乎只会从功能中获得故障结果。

我在C ++ DLL方面没有太多经验，但是其他几篇文章和帖子似乎在下面做到了。我还尝试在功能声明中传递外部函数的名称。我是错误地称其为DLL的问题吗？他们提供的应用程序（封闭式）能够正确使用它。

const
  SUSI_DLL = 'Susi4.dll';
  SUSI_STATUS_NOT_INITIALIZED = $FFFFFFFF;
  SUSI_STATUS_INITIALIZED = $FFFFFFFE;
  SUSI_STATUS_SUCCESS = $0;

function SusiLibInitialize: UInt32; stdcall; external SUSI_DLL;

{$R *.fmx}

procedure TForm1.btnTestClick(Sender: TObject);
begin
  var
  result := SusiLibInitialize;

  if result = SUSI_STATUS_INITIALIZED then
    ShowMessage('Init')
  else if result = SUSI_STATUS_SUCCESS then
    ShowMessage('Success')
  else if result = SUSI_STATUS_NOT_INITIALIZED then
    ShowMessage('Failure');
end;

原文

I'm attempting to use the SUSI api dll that was provided to me. The first step according to their documentation is to call SusiLibInitialize. When calling it I seem to only get the failure result from the function when I'm expected to get SUSI_STATUS_INITIALIZED as the result.

I do not have much experience with c++ dlls however several other articles and posts seems to do it as down below. I've also tried passing the name of the external function in the function declaration. Am I calling it incorrectly or could this be a issue with the DLL? The application they provided (closed sourced) is able to use it correctly.

const
  SUSI_DLL = 'Susi4.dll';
  SUSI_STATUS_NOT_INITIALIZED = $FFFFFFFF;
  SUSI_STATUS_INITIALIZED = $FFFFFFFE;
  SUSI_STATUS_SUCCESS = $0;

function SusiLibInitialize: UInt32; stdcall; external SUSI_DLL;

{$R *.fmx}

procedure TForm1.btnTestClick(Sender: TObject);
begin
  var
  result := SusiLibInitialize;

  if result = SUSI_STATUS_INITIALIZED then
    ShowMessage('Init')
  else if result = SUSI_STATUS_SUCCESS then
    ShowMessage('Success')
  else if result = SUSI_STATUS_NOT_INITIALIZED then
    ShowMessage('Failure');
end;