Perfect forward variadic arguments to lambda

Somehow I need to implement lazy evaluation with c++ variadic lambda. I am not quite sure whether the following code works correctly.

template <typename... ArgsT>
auto lazy_pack(ArgsT&& ... args) {

  auto T = [&](bool condition) {
    if(condition == false) return; 
    Foo v(std::forward<ArgsT>(args)...);
    v.do_work();
  };  

  return T;
}

The question is, how do I capture a given argument list and perfectly forward them to another templated object? The above example compiles but I am worried about if any dangling reference might happen. Another way is to capture arguments through copy and them and pass to the object:

template <typename... ArgsT>
auto lazy_pack(ArgsT&& ... args) {

  auto T = [=](bool condition) {
    if(condition == false) return; 
    Foo v(args...);
    v.do_work();
  };  

  return T;
}

---

I'm not sure I got what you are looking for, but maybe the following code can help you:

#include <tuple>
#include <functional>
#include <cstddef>
#include <utility>
#include <iostream>

struct Foo {
    Foo(int v, const char *s): val{v}, str{s} { }
    void do_work() { std::cout << val << " " << str << std::endl; }
    int val;
    const char *str;
};

template<std::size_t... I, typename... ArgsT>
auto lazy_pack(std::index_sequence<I...>, ArgsT&&... args) {
    return [tup{std::forward_as_tuple(std::forward<ArgsT>(args)...)}](bool condition) {
        if(condition == false) return;
        Foo v(std::get<I>(tup)...);
        v.do_work();
    };
}

template<typename... ArgsT>
auto lazy_pack(ArgsT&& ... args) {
    return lazy_pack(std::make_index_sequence<sizeof...(ArgsT)>(), std::forward<ArgsT>(args)...);
}

int main() {
    auto l = lazy_pack(42, "bar");
    l(false);
    l(true);
}

You need at least one more helper function to be able to unpack later the parameters forwarded to the first one.

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