Question

这个问题相当简单，我会误解一些微不足道的事情。但由于我几个小时都找不到解决方案，请让我在这里问一个问题。

我想要做的是使用Boost.Python将push_back容器类的std::vector<T>方法导出为一些固定类型T（比如double）到Python。

为此，我编写了以下代码：

typedef std::vector<double> vector_double;
class_<vector_double>("vector_double")
    .def("append", &vector_double::push_back);

但是这不会编译出现以下错误：

/Users/kenta/experiments/bisite/include/bisite/conservation_score.hpp:25:7: note:
      candidate constructor (the implicit default constructor) not viable:
      requires 0 arguments, but 1 was provided
/Users/kenta/experiments/bisite/src/bisite/pybisite.cpp:90:10: error: no
      matching member function for call to 'def'
        .def("append", &vector_double::push_back);
        ~^~~
/usr/local/include/boost/python/class.hpp:234:11: note: candidate template
      ignored: couldn't infer template argument 'F'
    self& def(char const* name, F f)
          ^
/usr/local/include/boost/python/class.hpp:224:11: note: candidate function
      template not viable: requires single argument 'visitor', but 2 arguments
      were provided
    self& def(def_visitor<Derived> const& visitor)

... and more candidates

类型演绎（？）似乎失败但我无法弄清楚原因。当我将vector_double定义为std::vector<double>的子类时，上面的代码已成功编译。但出于某种原因我不想这样做。

你能教我解决这个问题吗？

我正在使用带有-std=c++11选项的clang ++和Boost.Python v1.56.0。

谢谢。

Answer 1

在我看来，问题是，C ++ 11中有两个版本的push_back。尝试这样做

static_cast<void (vector_double::*)(const double&)>(&vector_double::push_back);

如何将vector <t> :: push_back导出到Python？</t>

1 个答案: