boost::python::object 的 C++ 赋值不起作用。为什么？

Question

根据 documentation，这里的赋值应该有效，但它没有：

#include <boost/python.hpp>
#include <iostream>

int main(int, char **) {
    using namespace boost::python;
    Py_Initialize();

    object test = object(2.05); //this works fine
    test = 3.05;                //compiler error

    std::cout << extract<double>(test) << std::endl;

    Py_Finalize();
    return 0;
}

编译器输出如下：

g++ -std=c++1y -I/usr/local/Cellar/python3/3.5.0/Frameworks/Python.framework/Versions/3.5/include/python3.5m -I/usr/local/Cellar/boost/1.59.0/include -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"Test.d" -MT"Test.d" -o "Test.o" "../Test.cpp"
../Test.cpp:9:10: error: no viable overloaded '='
    test = 3.05;
    ~~~~ ^ ~~~~
/usr/local/Cellar/boost/1.59.0/include/boost/python/object_core.hpp:241:9: note: candidate function (the implicit move assignment operator) not viable: no known conversion from 'double' to 'boost::python::api::object' for 1st argument
  class object : public object_base
        ^
/usr/local/Cellar/boost/1.59.0/include/boost/python/object_core.hpp:241:9: note: candidate function (the implicit copy assignment operator) not viable: no known conversion from 'double' to 'const boost::python::api::object' for 1st argument
  class object : public object_base
        ^
1 error generated.
make: *** [Test.o] Error 1

文档指出：

object msg = "%s is bigger than %s" % make_tuple(NAME,name);
Demonstrates that you can write the C++ equivalent of "format" % x,y,z in Python, which is useful since there's no easy way to do that in std C++.

但是，我无法让它工作。为什么我的 boost::python 不能自动转换双倍数？实际上，不只是双打，而是任何类型。

这是我的环境信息：

OSX Yosemite
LLVM 6.1.0
Python 3.5.0
提升 1.59.0
日食IDE

编辑：有效

int main(int, char **) {
    Py_Initialize();

    //object test = object(2.05); //this works fine
    //test = 3.05;                //compiler error

    str name = "test";
    str NAME = name.upper();
    object msg = "%s is bigger than %s" % make_tuple(NAME,name);

    std::cout<<std::string(extract<const char*>(msg))<<std::endl;

    Py_Finalize();
    return 0;
}

因此看起来您不能像 ForEveR 所建议的那样在不首先创建 boost::python 对象的情况下分配 C++ 类型。

Answer 1

因为object构造函数是explicit

  template <class T>
  explicit object(T const& x);

禁止使用像object = T();这样的隐式转换。您可以使用以下内容：

object test = object(2.05);

boost::python::object 的 C++ 赋值不起作用。为什么？

C++ assignment to boost::python::object doesn't work. Why?

c++

python

boost-python

编辑：有效