PyArg_ParseTupleAndKeywords 抛出警告：ISO C++ 禁止将字符串常量转换为 'char*' [-Wwrite-strings]

Question

搜索有关如何使用PyArg_ParseTupleAndKeywords我发现这些问题的示例：

1. 如何使用 PyArg_ParseTupleAndKeywords 解析带有可选参数和关键字的元组？
2. PyArg_ParseTupleAndKeywords 如何工作？

他们都使用类似的东西 static char* kwlist[] = {"a", "b", NULL}

static int PyClass_init(PyClass* self, PyObject* args, PyObject* kwargs) {
    char* path;
    char* regex;
    static char* kwlist[] = {"", "", NULL};

    if( !PyArg_ParseTupleAndKeywords( args, kwargs, "s|s", kwlist, &path, &regex ) ) {
        return -1;
    }

    // other code ...
    return 0;
}

用language = "c++"on编译setup.py它并用它构建它-std=c++11会抛出这个警告：

x86_64-linux-gnu-gcc -pthread -DNDEBUG -g -fwrapv -O0 -Wall -g -fstack-protector-strong -Wformat -Werror=format-security -Wdate-time -D_FORTIFY_SOURCE=2 -fPIC -Isource -I/usr/include/python3.6m -c source/test.cpp -o build/temp.linux-x86_64-3.6/source/test.o -O0 -g -ggdb -std=c++11 -fstack-protector-all
source/test.cpp: In function ‘int PyClass_init(PyClass*, PyObject*, PyObject*)’:
source/test.cpp:41:42: warning: ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]
   static char* kwlist[] = {"a", "b", NULL};
                                          ^

搜索此错误时，我发现了这个问题Why is conversion from string constant to 'char*' 在 C 中有效，但在 C++ 中无效，需要修复，但将修复应用为static char* const kwlist[] = {"a", "b", NULL};保持/升级引入错误的警告：

source/test.cpp:41:50: warning: ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]
     static char* const kwlist[] = {"a", "b", NULL};
                                                  ^
source/test.cpp:41:50: warning: ISO C++ forbids converting a string constant to ‘char*’ [-Wwrite-strings]
source/test.cpp:43:89: error: invalid conversion from ‘char* const*’ to ‘char**’ [-fpermissive]
     if( !PyArg_ParseTupleAndKeywords( args, kwargs, "s|s", kwlist, &path, &regex ) ) {
                                                                                         ^
In file included from /usr/include/python3.6m/Python.h:117:0,
                 from source/test.cpp:3:
/usr/include/python3.6m/modsupport.h:17:41: note:   initializing argument 4 of ‘int _PyArg_ParseTupleAndKeywords_SizeT(PyObject*, PyObject*, const char*, char**, ...)’
 #define PyArg_ParseTupleAndKeywords     _PyArg_ParseTupleAndKeywords_SizeT
                                         ^
/usr/include/python3.6m/modsupport.h:17:41: note: in definition of macro ‘PyArg_ParseTupleAndKeywords’
 #define PyArg_ParseTupleAndKeywords     _PyArg_ParseTupleAndKeywords_SizeT
                                         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
error: command 'x86_64-linux-gnu-gcc' failed with exit status 1

在与 Python C API 要求兼容的同时，如何使用static char* kwlist[] = {"a", "b", NULL}与 with等效的构造来消除警告C++ 11？

---

更新

建议后我尝试static const char* kwlist[] = {"a", "b", NULL}但PyArg_ParseTupleAndKeywords不接受：

source/test.cpp:43:89: error: invalid conversion from ‘const char**’ to ‘char**’ [-fpermissive]
     if( !PyArg_ParseTupleAndKeywords( args, kwargs, "s|s", kwlist, &filepath, &rawregex ) ) {
                                                                                         ^
In file included from /usr/include/python3.6m/Python.h:117:0,
                 from source/test.cpp:3:
/usr/include/python3.6m/modsupport.h:17:41: note:   initializing argument 4 of ‘int _PyArg_ParseTupleAndKeywords_SizeT(PyObject*, PyObject*, const char*, char**, ...)’
 #define PyArg_ParseTupleAndKeywords     _PyArg_ParseTupleAndKeywords_SizeT
                                         ^
/usr/include/python3.6m/modsupport.h:17:41: note: in definition of macro ‘PyArg_ParseTupleAndKeywords’
 #define PyArg_ParseTupleAndKeywords     _PyArg_ParseTupleAndKeywords_SizeT
                                         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
error: command 'x86_64-linux-gnu-gcc' failed with exit status 1

Answer 1

您复制的示例代码可能是 C，其中警告没有被认真考虑（gcc在此类代码上没有报告错误，g++即使没有警告选项也会弹出警告）。

这里的解决方法是添加这样的const关键字：

static const char* kwlist[] = {"a", "b", NULL};

const您尝试的关键字不是应用于值而是应用于指针。

然后使函数接受char **使用const_cast：

const_cast<char **>(kwlist)

像这样：

PyArg_ParseTupleAndKeywords( args, kwargs, "s|s", const_cast<char **>(kwlist), &path, &regex ) ) 

该函数不会通过隐式“契约”更改值（因为它是 python API，它不会骗你），所以在这种情况下，转换为 const 很好）

PS：这个其他答案给出的替代方案也有效。请注意，使用空的可变字符串并不比转换为常量更安全，就好像被调用的函数决定写入空字符串一样，它也会调用未定义的行为。