这是在定义一个 lambda 函数并同时将函数指针分配给一个值吗？

Question

Still many C++ codes are so difficult for me to understand..
Below is a code snippet from dlib ( http://dlib.net file : dlib/external/pybind11/include/pybind11/pybind11.h)
It's a member function definition of class cpp_function and I didn't try to understand the code(no time to do that..that's sad..). I can't understand the syntax in the line I put *** this line! comment at below. I understand the lambda function(unnamed function), so is it assigning a function pointer to rec->impl , the function taking function_call &call as argument and returning handle ? So, it looks like defining a function and at the same time assigning the function pointer to a variable. Having asked it, it looks so now.. Please someone confirm this.

    void initialize(Func &&f, Return (*)(Args...), const Extra&... extra) {
        using namespace detail;

        struct capture { remove_reference_t<Func> f; };

...

        rec->impl = [](function_call &call) -> handle {    // <=== *** this line!
            cast_in args_converter;

            /* Try to cast the function arguments into the C++ domain */
            if (!args_converter.load_args(call))
                return PYBIND11_TRY_NEXT_OVERLOAD;

            /* Invoke call policy pre-call hook */
            process_attributes<Extra...>::precall(call);

            /* Get a pointer to the capture object */
            auto data = (sizeof(capture) <= sizeof(call.func.data)
                         ? &call.func.data : call.func.data[0]);
            capture *cap = const_cast<capture *>(reinterpret_cast<const capture *>(data));

            /* Override policy for rvalues -- usually to enforce rvp::move on an rvalue */
            const auto policy = return_value_policy_override<Return>::policy(call.func.policy);

            /* Function scope guard -- defaults to the compile-to-nothing `void_type` */
            using Guard = extract_guard_t<Extra...>;

            /* Perform the function call */
            handle result = cast_out::cast(
                std::move(args_converter).template call<Return, Guard>(cap->f), policy, call.parent);

            /* Invoke call policy post-call hook */
            process_attributes<Extra...>::postcall(call, result);

            return result;
        };

...
        using FunctionType = Return (*)(Args...);
        constexpr bool is_function_ptr =
            std::is_convertible<Func, FunctionType>::value &&
            sizeof(capture) == sizeof(void *);
        if (is_function_ptr) {
            rec->is_stateless = true;
            rec->data[1] = const_cast<void *>(reinterpret_cast<const void *>(&typeid(FunctionType)));
        }
    }

Answer 1

rec->impl = [](function_call &call) -> handle

creates a lambda which takes one argument of type function_call and returns a handle , then assigns it to rec->impl .

As lambdas are basically unnamed structs, they also have unnamed types. Since rec->impl obviously exists already and is thus not templatized on the lambda type, the lambda gets converted to some other type during the assignment. (Note: there could however be a templatized and overloaded operator= here)

Typically such types which can take lambdas are either std::function or function pointers as stateless lambdas can be converted to function pointers.