在 Python 中使用回调时出错

2022-01-12 00:00:00 python error-handling char callback c++

我正在开发一个应该在 Python 中使用的 dll.我有一个回调函数来发送我的参数(在单独的标头中定义):

I am developing a dll that should be used in Python. I have a callback function to send my parameters (defined in a separate header):

typedef int(*call_nBest)(char **OutList, float* confList, int nB);

所以，我就是这样使用这个回调的:

So, I'm using this callback in this way:

#define TEXT_BUFFER_MAX_SIZE 50
call_nBest nBestList;
void Xfunction(const char* aLineThatWillBeConvertedInAList){
    char **results;
    float *confidences;
    confidences=new float[nBest];
    results=new char*[nBest];
    for(int i=0; i<nBest; i++) results[i]=new char[TEXT_BUFFER_MAX_SIZE];

    MakeLine2List(aLineThatWillBeConvertedInAList,results,confidences); 

    /*At this function I am having the error :(*/
    nBestList(results,confidences,nBest); // Passing the values to my callback

    for(int i=0; i<nBest; i++) delete [] results[i];
    delete [] confidences;
    delete [] results;

}

我是这样导出的:

__declspec(dllexport) int ResultCallback(call_nBest theList){
    nBestList = theList;
    return(0);
}

我首先以这种方式在另一个 C++ 应用程序中测试了我的回调:

I tested my callback first in another C++ application in this way:

int MyCallback(char **OutLi, float* confLi, int nB){
    printf("
 The nB results: %d 
",nB);
    for(int n=0; n<nB; n++){
        std::cout << *(confLi+n) << "	" << OutLi[n] << "
";
    }
    return(0);
}

在main()中我是这样给回调的:

In main() I give the callback in this way:

ResultCallback(MyCallback);

而且效果很好.但我不知道如何使它适应 Python.我试过这个:

and it works pretty well. But I don't have any idea how to adapt this to Python. I have tried this:

注意:我改变了最后一种方式，因为我解决了一些错误，但我仍然收到错误.这是我加载 myDLL

Note: I have changed the last way, because I resolved some mistakes, but I'm still getting an error. This is the current way of how I am loading myDLL

from ctypes import *
def callbackU(OutList,ConList,nB):
    for i in range(nB):
        print(OutList[i][0:50]) #I don't know how to print the values
return 0

myDLL = cdll.LoadLibrary("MyLibrary.dll")

calling = CFUNCTYPE(c_int,POINTER(POINTER(c_char)),POINTER(c_float),c_int)
theCall= calling(callbackU)
myDLL.ResultCallback(theCall)

myDLL.StartProcess(); #In this process the given callback will be invoqued

错误

现在我有这个错误:

ERROR

And now I have this error:

未处理的异常:System.AccessViolationException:试图读或写受保护的内存.这通常表明其他内存已损坏.在 Xfunction(SByte*aLineThatWillBeConvertedInAList)

Unhandled Exception: System.AccessViolationException: Attempted to read or write protected memory. This is often an indication that other memory is corrupt. at Xfunction(SByte* aLineThatWillBeConvertedInAList)

问题事件名称:APPCRASH
应用程序名称:python.exe
应用程序版本:0.0.0.0
申请时间戳:54f9ed12
故障模块名称:MSVCR100.dll
故障模块版本:10.0.40219.325
故障模块时间戳:10.0.40219.325
异常代码:c0000005
异常偏移:00001ed7
操作系统版本:6.3.9600.2.0.0.256.4
区域设置 ID:1033
附加信息 1:5861
附加信息2:5861822e1919d7c014bbb064c64908b2
附加信息 3:a10f
附加信息4:a10ff7d2bb2516fdc753f9c34fc3b069

Problem Event Name: APPCRASH
Application Name: python.exe
Application Version: 0.0.0.0
Application Timestamp: 54f9ed12
Fault Module Name: MSVCR100.dll
Fault Module Version: 10.0.40219.325
Fault Module Timestamp: 10.0.40219.325
Exception Code: c0000005
Exception Offset: 00001ed7
OS Version: 6.3.9600.2.0.0.256.4
Locale ID: 1033
Additional Information 1: 5861
Additional Information 2: 5861822e1919d7c014bbb064c64908b2
Additional Information 3: a10f
Additional Information 4: a10ff7d2bb2516fdc753f9c34fc3b069

我已经做过并且几乎是我想要的事情:

首先我为这个更改了回调 Python 函数:

Things that I've done and are almost what I want:

First I changed the callback Python function for this one:

def callbackU(OutList,ConList,nB):
    for i in range(nB):
        print(i)
return 0

一切正常，我可以在控制台中看到这一点(在这种情况下，nB 是 10):

All works with no error and I can see this in the Console (in this case nB was 10):

0
1
...
9

其次，我把功能改成了这个:

Second, I changed the function as this one:

def callbackU(OutList,ConList,nB):
    for i in range(nB):
        print (cast(OutList,c_char_p))
return 0

而且，哦，令人惊讶的是，这只会打印列表的第一个单词(nB 次)

and, oh surprise this prints only the first word of the list (nB times)

推荐答案

你想要这样的东西吗?

def callbackU(OutList, ConList, nB):
    for i in range(nB):
        print("{}	{}".format(ConList[i], cast(OutList[i], c_char_p)))
    return 0

据我了解，您只是想将 Python callbackU 函数的输出与 C++ MyCallback 函数相匹配.

From what I understand you're just trying to match the output of your Python callbackU function with your C++ MyCallback function.

Python 有多种字符串格式化功能，一开始可能会让人困惑，但它向 printf 致敬 字符串格式化.

Python has a variety of string formatting functionality that can be confusing at first, but pays homage to printf string formatting.

由于 OutList 的类型为 LP_LP_c_char(指向 c_char 的指针的指针，vs "NULL 终止的 char *"c_char_p)，我们最好把它变成一个原生的 Python 数据类型，像这样:

Since OutList has type LP_LP_c_char (pointer to pointer of c_char, vs "NULL terminated char *" c_char_p), we'd best turn it into a native Python data type like so:

def callbackU(OutList, ConList, nB):
    for i in range(nB):
        out_list_item = cast(OutList[i], c_char_p).value
        print("{}	{}".format(ConList[i], out_list_item))
    return 0

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