Python 和 ctypes：如何将“指针到指针”正确传递到 DLL 中？
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Python and ctypes: how to correctly pass "pointer-to-pointer" into DLL?
提问by grigoryvp
I have a DLL that allocates memory and returns it. Function in DLL is like this:
我有一个分配内存并返回它的 DLL。DLL 中的函数是这样的：
void Foo( unsigned char** ppMem, int* pSize )
{
* pSize = 4;
* ppMem = malloc( * pSize );
for( int i = 0; i < * pSize; i ++ ) (* ppMem)[ i ] = i;
}
Also, i have a python code that access this function from my DLL:
另外，我有一个从我的 DLL 访问这个函数的 python 代码：
from ctypes import *
Foo = windll.mydll.Foo
Foo.argtypes = [ POINTER( POINTER( c_ubyte ) ), POINTER( c_int ) ]
mem = POINTER( c_ubyte )()
size = c_int( 0 )
Foo( byref( mem ), byref( size ) ]
print size, mem[ 0 ], mem[ 1 ], mem[ 2 ], mem[ 3 ]
I'm expecting that printwill show "4 0 1 2 3" but it shows "4 221 221 221 221" O_O. Any hints what i'm doing wrong?
我期待它print会显示“4 0 1 2 3”，但它会显示“4 221 221 221 221”O_O。任何提示我做错了什么？
采纳答案by Mark Tolonen
Post actual code. The C/C++ code doesn't compile as either C or C++. The Python code has syntax errors (] ending function call Foo). The code below works. The main issue after fixing syntax and compiler errors was declaring the function __stdcallso windllcould be used in the Python code. The other option is to use __cdecl(normally the default) and use cdllinstead of windllin the Python code.
发布实际代码。C/C++ 代码不会编译为 C 或 C++。Python 代码有语法错误 (] 结束函数调用 Foo)。下面的代码有效。修复语法和编译器错误后的主要问题是声明函数，__stdcall以便windll可以在 Python 代码中使用。另一个选项是在 Python 代码中使用__cdecl（通常是默认值）和使用cdll代替windll。
mydll.c (cl /W4 /LD mydll.c)
mydll.c (cl /W4 /LD mydll.c)
#include <stdlib.h>
__declspec(dllexport) void __stdcall Foo(unsigned char** ppMem, int* pSize)
{
char i;
*pSize = 4;
*ppMem = malloc(*pSize);
for(i = 0; i < *pSize; i++)
(*ppMem)[i] = i;
}
demo.py
演示文件
from ctypes import *
Foo = windll.mydll.Foo
Foo.argtypes = [POINTER(POINTER(c_ubyte)),POINTER(c_int)]
mem = POINTER(c_ubyte)()
size = c_int(0)
Foo(byref(mem),byref(size))
print size.value,mem[0],mem[1],mem[2],mem[3]
Output
输出
4 0 1 2 3
