问题描述
我想用 C++17 std::variant 替换 boost::variant 并去掉 boost::recursive_wrappercode>，在下面的代码中完全消除对boost的依赖.我该怎么做?
I'd like to replace boost::variants with C++17 std::variant and get rid of boost::recursive_wrapper, to remove dependency on boost completely in following code. How may I do that?
#include <boost/variant.hpp>
#include <type_traits>
using v = boost::variant<int, boost::recursive_wrapper<struct s> >;
struct s
{
v val;
};
template<template <typename...> class R, typename T, typename ... Ts>
auto reduce(T t, Ts ... /*ts*/)
{
return R<T, Ts...>{t};
}
template<typename T, typename F>
T adapt(F f)
{
static_assert(std::is_convertible_v<F, T>, "");
return f;
}
int main()
{
int val1 = 42;
s val2;
auto val3 = adapt<v>(reduce<boost::variant>(val1, val2));
}
有两个通用函数:第一个函数reduce在运行时选择返回哪个参数(这里为了简洁起见它只返回第一个参数)，第二个函数adapt转换一个值类型 F 到类型 T 的值.
There are two generic functions: first function reduce chooses at runtime which argument to return (here it just returns first argument for brevity), second function adapt converts a value of type F to a value of type T.
在这个例子中，reduce 返回一个 boost::variant 类型的对象，然后它被转换成一个 boost:: 类型的对象变体<int, boost::recursive_wrapper<s>>.
In this example reduce returns an object of type boost::variant<int, s> which is then converted to an object of type boost::variant<int, boost::recursive_wrapper<s> >.
推荐答案
boost::variant 将堆分配以便将自身的一部分递归地定义为自身.(它也会在其他一些情况下堆分配，不知道有多少)
boost::variant will heap allocate in order to have part of itself be recursively defined as itself. (It will also heap allocate in a number of other situations, uncertain how many)
std::variant 不会.std::variant 拒绝堆分配.
std::variant will not. std::variant refuses to heap allocate.
在没有动态分配的情况下，没有办法真正拥有包含自身可能变体的结构，因为如果静态声明，这样的结构很容易显示为无限大.(您可以通过 N 个不相同的递归来编码整数 N:没有固定大小的缓冲区可以容纳无限量的信息.)
There is no way to actually have a structure containing a possible variant of itself without a dynamic allocation, as such a structure can easily be shown to be infinite in size if statically declared. (You can encode the integer N by having N recursions of not-the-same: no fixed size buffer can hold an infinite amount of information.)
因此，等效的 std::variant 存储了一个智能指针，它是自身递归实例的某种占位符.
As such, the equivalent std::variant stores a smart pointer of some kind placeholder of a recursive instance of itself.
这可能有效:
struct s;
using v = std::variant< int, std::unique_ptr<s> >;
struct s
{
v val;
~s();
};
inline s::~s() = default;
如果失败，请尝试:
struct destroy_s;
struct s;
using v = std::variant<int, std::unique_ptr<s, destroy_s> >;
struct s
{
v val;
~s();
};
struct destroy_s {
void operator()(s* ptr){ delete ptr; }
};
inline s::~s() = default;
这确实意味着客户端代码必须有意识地与 unique_ptr 而不是 struct s 直接交互.
It does mean that client code has to knowingly interact with the unique_ptr<s> and not the struct s directly.
如果你想支持复制语义，你必须编写一个执行复制的 value_ptr，并赋予它等效的 struct copy_s; 来实现该复制.
If you want to support copy semantics, you'll have to write a value_ptr that does copies, and give it the equivalent of struct copy_s; to implement that copy.
template<class T>
struct default_copier {
// a copier must handle a null T const* in and return null:
T* operator()(T const* tin)const {
if (!tin) return nullptr;
return new T(*tin);
}
void operator()(void* dest, T const* tin)const {
if (!tin) return;
return new(dest) T(*tin);
}
};
template<class T, class Copier=default_copier<T>, class Deleter=std::default_delete<T>,
class Base=std::unique_ptr<T, Deleter>
>
struct value_ptr:Base, private Copier {
using copier_type=Copier;
// also typedefs from unique_ptr
using Base::Base;
value_ptr( T const& t ):
Base( std::make_unique<T>(t) ),
Copier()
{}
value_ptr( T && t ):
Base( std::make_unique<T>(std::move(t)) ),
Copier()
{}
// almost-never-empty:
value_ptr():
Base( std::make_unique<T>() ),
Copier()
{}
value_ptr( Base b, Copier c={} ):
Base(std::move(b)),
Copier(std::move(c))
{}
Copier const& get_copier() const {
return *this;
}
value_ptr clone() const {
return {
Base(
get_copier()(this->get()),
this->get_deleter()
),
get_copier()
};
}
value_ptr(value_ptr&&)=default;
value_ptr& operator=(value_ptr&&)=default;
value_ptr(value_ptr const& o):value_ptr(o.clone()) {}
value_ptr& operator=(value_ptr const&o) {
if (o && *this) {
// if we are both non-null, assign contents:
**this = *o;
} else {
// otherwise, assign a clone (which could itself be null):
*this = o.clone();
}
return *this;
}
value_ptr& operator=( T const& t ) {
if (*this) {
**this = t;
} else {
*this = value_ptr(t);
}
return *this;
}
value_ptr& operator=( T && t ) {
if (*this) {
**this = std::move(t);
} else {
*this = value_ptr(std::move(t));
}
return *this;
}
T& get() { return **this; }
T const& get() const { return **this; }
T* get_pointer() {
if (!*this) return nullptr;
return std::addressof(get());
}
T const* get_pointer() const {
if (!*this) return nullptr;
return std::addressof(get());
}
// operator-> from unique_ptr
};
template<class T, class...Args>
value_ptr<T> make_value_ptr( Args&&... args ) {
return {std::make_unique<T>(std::forward<Args>(args)...)};
}
value_ptr 的实例.
这篇关于将 std::variant 与递归一起使用，而不使用 boost::recursive_wrapper的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！