Python swig c++;通过引用返回无符号字符*
我使用SWIG将C++函数包到Python中:Python swig c++;通过引用返回无符号字符*,python,c++,swig,unsigned-char,Python,C++,Swig,Unsigned Char,我使用SWIG将C++函数包到Python中: int getBytes(unsigned char *& result, int length) 我学习了如何包装通过引用返回的char*函数: %typemap(in,numinputs=0) char*& (char* tmp) %{ $1 = &tmp; %} %typemap(argout) char*& (PyObject* obj) %{ obj = PyUnicode_FromSt
int getBytes(unsigned char *& result, int length)
我学习了如何包装通过引用返回的char*函数:
%typemap(in,numinputs=0) char*& (char* tmp) %{
$1 = &tmp;
%}
%typemap(argout) char*& (PyObject* obj) %{
obj = PyUnicode_FromString(*$1);
$result = SWIG_Python_AppendOutput($result,obj);
%}
%typemap(freearg) char*& %{
free(*$1);
%}
我试图将其应用于无符号字符*,但失败,并得到错误信息,即SWIG\u Python\u AppendOutput
无法将无符号字符**转换为有符号字符**。我搜索了python文档,没有找到可以将无符号字符**转换为无符号字符**的函数。
有人能帮忙吗?谢谢 我很久以前就解决了这个问题,下面是我的解决方案:
%typemap(in, numinputs= 0) (unsigned char *&result, int &length) (unsigned char *temp, int len) %{
$1 = &temp;
$2 = &len;
%}
%typemap(argout) (unsigned char *&result, int &length) %{
PyObject* arg5 = 0;
arg5 = SWIG_FromCharPtrAndSize((const char *)(*$1), *$2);
PyObject* temp = NULL;
temp = $result;
$result = PyList_New(1);
PyList_SetItem($result, 0, temp);
PyList_Append($result, (PyObject*)arg5);//unsigned char *&
PyList_Append($result, PyInt_FromLong(*$2));
Py_DECREF(temp);
%}