包装的 C++ 函数的输入信号不会改变 SWIG [英] Input sting for wrapped C++ function doesn't changing SWIG

问题描述

我为 C++ 库制作 Python 包装器.

I make Python wrapper for C++ library.

mylib.i:

%module mylib
%include <std_string.i>

%{
    #include "mylib.h"

%}
%apply const std::string & {std::string &};
%apply std::string & {std::string &};

int getResult(const std::string& path, std::string& result);

mylib.h:

#pragma once

#include <string>

myEnum {foo=0, bar};
myEnum getResult(const std::string& path, std::string& result);

使用以下命令生成 _mylib.so 后:

After generating _mylib.so with following command:

g++ -fPIC -Wall -Wextra -shared mylib_wrap.cxx -o _mylib.so -L. -lmylib -I/usr/include/python2.7/ -lpython2.7

我下一步:

LD_LIBRARY_PATH=. python Python 2.7.2 (default, Dec  6 2016, 10:43:39) 
[GCC 4.8.4] on linux4
Type "help", "copyright", "credits" or "license" for more information.
>>> import _mylib
>>> result= ""
>>> x = _mylib.getResult("somePath",result)

执行我的函数后，x 返回方法的正确响应.我的函数也有控制台输出.但结果字符串不会改变.如果我用some text"初始化结果字符串，并再次调用我的函数，print result 返回some text".我做错了什么?

After execution of my function, x returns right respond of method. Also I have console output from my function. But result string doesn't changing. If I init result string with "some text", and call my function again, print result return "some text". What I'm doing wrong?

推荐答案

让这个在你的界面中工作的最简单的方法是使用 %inline 创建一个重载来返回结果:

The simplest way to get this to work inside your interface is to use %inline to create an overload that returns the result instead:

%module mylib
%include <std_string.i>

%{
    #include "mylib.h"
%}

%inline %{
    int getResult(const std::string& path) {
        std::string temp;
        const int ret = getResult(path, temp);
        if (ret != good) abort(); // TODO exceptions
        return temp;
    }
%}

您甚至不需要向 SWIG 展示 getResult 的真实版本.

You don't even need to show SWIG the real version of getResult.

还有其他选项(例如，使用 numinputs=0 的输入类型映射和 argout 类型映射来修改返回值)，但它们更复杂，通常对其他语言的移植性较差.

There are other options (e.g. use an input typemap with numinputs=0 and an argout typemap to modify the return value), but they're more complicated and generally less portable to other languages.

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