python - 简化在Pybind11中为C++模板类生成包装器类的步骤:模板声明不能出现在 block 作用域中

Question

我试图简化在Pybind11中为C++模板类生成包装器类的过程。这是一个显示问题的最小示例(以下this回答):

#include <pybind11/pybind11.h>
#include <iostream>

namespace py = pybind11;

template<class T>
class Foo {
public:
    Foo(T bar) : bar_(bar) {}
    void print() {
        std::cout << "Type id: " << typeid(T).name() << '\n';
    }
private:
    T bar_;
};

PYBIND11_MODULE(example, m) {
    template<typename T>
    void declare_foo(py::module &m, std::string &typestr) {
        using Class = Foo<T>;
        std::string pyclass_name = std::string("Foo") + typestr;
        py::class_<Class>(m, pyclass_name.c_str())
            .def(py::init< T >())
            .def("print", &Class::print);
    }
    declare_foo<int>(m, "Int");
    declare_foo<double>(m, "Double");
    # More similar declarations follow here...
}

当我用以下命令编译时:

g++ -O3 -Wall -shared -std=c++17 -fPIC `python3 -m pybind11 --includes` example.cpp -o example`python3-config --extension-suffix`

我得到错误:

example.cpp: In function ‘void pybind11_init_example(pybind11::module&)’:
example.cpp:18:5: error: a template declaration cannot appear at block scope
   18 |     template<typename T>
      |     ^~~~~~~~

Answer 1

像错误提示一样，您不能在块范围内使用模板声明(显然您在https://en.cppreference.com/w/cpp/language/scope中)。只需将其移到外部并通过const引用(或值)捕获字符串参数即可。

将代码更改为

template<typename T>
void declare_foo(py::module &m, const std::string &typestr) {
    using Class = Foo<T>;
    std::string pyclass_name = std::string("Foo") + typestr;
    py::class_<Class>(m, pyclass_name.c_str())
        .def(py::init< T >())
        .def("print", &Class::print);
}

PYBIND11_MODULE(example, m) {
    declare_foo<int>(m, "Int");
    declare_foo<double>(m, "Double");
}

作品。

作为旁注，您还应该

#include <string>

不建议依赖可传递包含。