C++20 中的完美捕捉
template <typename ... Args>
auto f(Args&& ... args){
return [... args = std::forward<Args>(args)]{
// use args
};
}
C++17 和 C++14 解决方法
在 C++17 中，我们可以使用元组的解决方法：
template <typename ... Args>
auto f(Args&& ... args){
return [args = std::make_tuple(std::forward<Args>(args) ...)]()mutable{
return std::apply([](auto&& ... args){
// use args
}, std::move(args));
};
}
不幸的是 std::apply 是 C++17，在 C++14 中你可以自己实现它或者用 boost::hana 做类似的事情：
namespace hana = boost::hana;
template <typename ... Args>
auto f(Args&& ... args){
return [args = hana::make_tuple(std::forward<Args>(args) ...)]()mutable{
return hana::unpack(std::move(args), [](auto&& ... args){
// use args
});
};
}
通过函数capture_call 简化解决方法可能很有用：
#include <tuple>
// Capture args and add them as additional arguments
template <typename Lambda, typename ... Args>
auto capture_call(Lambda&& lambda, Args&& ... args){
return [
lambda = std::forward<Lambda>(lambda),
capture_args = std::make_tuple(std::forward<Args>(args) ...)
](auto&& ... original_args)mutable{
return std::apply([&lambda](auto&& ... args){
lambda(std::forward<decltype(args)>(args) ...);
}, std::tuple_cat(
std::forward_as_tuple(original_args ...),
std::apply([](auto&& ... args){
return std::forward_as_tuple< Args ... >(
std::move(args) ...);
}, std::move(capture_args))
));
};
}
像这样使用它：
#include <iostream>
// returns a callable object without parameters
template <typename ... Args>
auto f1(Args&& ... args){
return capture_call([](auto&& ... args){
// args are perfect captured here
// print captured args via C++17 fold expression
(std::cout << ... << args) << '\n';
}, std::forward<Args>(args) ...);
}
// returns a callable object with two int parameters
template <typename ... Args>
auto f2(Args&& ... args){
return capture_call([](int param1, int param2, auto&& ... args){
// args are perfect captured here
std::cout << param1 << param2;
(std::cout << ... << args) << '\n';
}, std::forward<Args>(args) ...);
}
int main(){
f1(1, 2, 3)(); // Call lambda without arguments
f2(3, 4, 5)(1, 2); // Call lambda with 2 int arguments
}
这是capture_call的C++14实现：
#include <tuple>
// Implementation detail of a simplified std::apply from C++17
template < typename F, typename Tuple, std::size_t ... I >
constexpr decltype(auto)
apply_impl(F&& f, Tuple&& t, std::index_sequence< I ... >){
return static_cast< F&& >(f)(std::get< I >(static_cast< Tuple&& >(t)) ...);
}
// Implementation of a simplified std::apply from C++17
template < typename F, typename Tuple >
constexpr decltype(auto) apply(F&& f, Tuple&& t){
return apply_impl(
static_cast< F&& >(f), static_cast< Tuple&& >(t),
std::make_index_sequence< std::tuple_size<
std::remove_reference_t< Tuple > >::value >{});
}
// Capture args and add them as additional arguments
template <typename Lambda, typename ... Args>
auto capture_call(Lambda&& lambda, Args&& ... args){
return [
lambda = std::forward<Lambda>(lambda),
capture_args = std::make_tuple(std::forward<Args>(args) ...)
](auto&& ... original_args)mutable{
return ::apply([&lambda](auto&& ... args){
lambda(std::forward<decltype(args)>(args) ...);
}, std::tuple_cat(
std::forward_as_tuple(original_args ...),
::apply([](auto&& ... args){
return std::forward_as_tuple< Args ... >(
std::move(args) ...);
}, std::move(capture_args))
));
};
}
capture_call 按值捕获变量。完美意味着尽可能使用移动构造函数。下面是一个 C++17 代码示例，以便更好地理解：
#include <tuple>
#include <iostream>
#include <boost/type_index.hpp>
// Capture args and add them as additional arguments
template <typename Lambda, typename ... Args>
auto capture_call(Lambda&& lambda, Args&& ... args){
return [
lambda = std::forward<Lambda>(lambda),
capture_args = std::make_tuple(std::forward<Args>(args) ...)
](auto&& ... original_args)mutable{
return std::apply([&lambda](auto&& ... args){
lambda(std::forward<decltype(args)>(args) ...);
}, std::tuple_cat(
std::forward_as_tuple(original_args ...),
std::apply([](auto&& ... args){
return std::forward_as_tuple< Args ... >(
std::move(args) ...);
}, std::move(capture_args))
));
};
}
struct A{
A(){
std::cout << " A::A()\n";
}
A(A const&){
std::cout << " A::A(A const&)\n";
}
A(A&&){
std::cout << " A::A(A&&)\n";
}
~A(){
std::cout << " A::~A()\n";
}
};
int main(){
using boost::typeindex::type_id_with_cvr;
A a;
std::cout << "create object end\n\n";
[b = a]{
std::cout << " type of the capture value: "
<< type_id_with_cvr<decltype(b)>().pretty_name()
<< "\n";
}();
std::cout << "value capture end\n\n";
[&b = a]{
std::cout << " type of the capture value: "
<< type_id_with_cvr<decltype(b)>().pretty_name()
<< "\n";
}();
std::cout << "reference capture end\n\n";
[b = std::move(a)]{
std::cout << " type of the capture value: "
<< type_id_with_cvr<decltype(b)>().pretty_name()
<< "\n";
}();
std::cout << "perfect capture end\n\n";
[b = std::move(a)]()mutable{
std::cout << " type of the capture value: "
<< type_id_with_cvr<decltype(b)>().pretty_name()
<< "\n";
}();
std::cout << "perfect capture mutable lambda end\n\n";
capture_call([](auto&& b){
std::cout << " type of the capture value: "
<< type_id_with_cvr<decltype(b)>().pretty_name()
<< "\n";
}, std::move(a))();
std::cout << "capture_call perfect capture end\n\n";
}
输出：
A::A()
create object end
A::A(A const&)
type of the capture value: A const
A::~A()
value capture end
type of the capture value: A&
reference capture end
A::A(A&&)
type of the capture value: A const
A::~A()
perfect capture end
A::A(A&&)
type of the capture value: A
A::~A()
perfect capture mutable lambda end
A::A(A&&)
type of the capture value: A&&
A::~A()
capture_call perfect capture end
A::~A()
捕获值的类型在capture_call 版本中包含&&，因为我们必须通过引用访问内部元组中的值，而支持的语言捕获支持直接访问该值。