成员变量的结构化绑定 [英] structured binding of member variables

问题描述

在Visual Studio 2017中，打开/std:c ++ 17 后，我可以做

In visual studio 2017, after turning on /std:c++17, I can do

auto [d1, d2] = func_return_tuple();

，其中 func_return_tuple()返回两个双精度值的元组.

, where func_return_tuple() returns a tuple of two double value.

但是，如果类A有两个成员变量 d1 _ 和 d2 _ ，我就不能这样做

However, if class A has two member variable d1_ and d2_, I can't do

A a;
auto [a.d1_, a.d2_] = func_return_tuple();

有没有出路?

当然

std::tie(a.d1_, a.d2_) = func_return_tuple();

始终有效.

推荐答案

有没有出路?

基本上，不.结构化绑定仅与将对象分解为组成元素有关.它总是引入 n 个新名称-每个元素一个.您不能通过结构化绑定进行分配，也不能使用它来重新绑定现有名称或类似名称.

Basically, no. Structured bindings is only about decomposing an object into constituent elements. It always introduces n new names - one for each element. You cannot assign through a structured binding, nor can you use it to rebind existing names, or anything like that.

您的选择基本上是:

std::tie(a.d1_, a.d2_) = func_return_tuple();

---

或者，假设 A 是包含这两个成员的聚合，例如:

---

Or, assuming A is an aggregate containing those two members, something like:

template <typename T>
struct list_init_t {
    template <typename... Args>
    T operator()(Args&&... args) const {
        return T{std::forward<Args>(args)...};
    }
};

template <typename T>
inline constexpr list_init_t<T> list_init;

用作:

a = std::apply(list_init<A>, func_return_tuple());

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