结构化绑定和tie（） [英] Structured binding and tie()

问题描述

给出以下声明：

int a[3] {10,20,30};
std::tuple<int,int,int> b {11,22,33};

我可以使用结构化绑定声明来解码 a 和 b ：

I can use structured binding declarations to decode a and b:

auto [x1,y1,z1] = a;
auto [x2,y2,z2] = b;

但是如果 x1 ， y1 等已经存在，该怎么办？

But if x1, y1, etc. already exist, what do I do?

std::tie(x1,y1,z1) = a;  // ERROR
std::tie(x2,y2,z2) = b;  // OK

这适用于 b ，但不适用于 a 。是否有类似的简单构造适用于 a ，还是我必须获取 a [0] ，<$分别是c $ c> a [1] 和 a [2] ？

This works for b but not for a. Is there a similar simple construct that works for a, or do I have to fetch a[0], a[1] and a[2] separately?

推荐答案

不是。

结构化绑定具有用于处理数组和某些其他类型的特定语言规则。 tie（）特别是 tuple< T& ...> ，并且只能从另一个 tuple< U& ...> 。

Structured bindings has specific language rules to handle arrays and certain other types. tie() is specifically a tuple<T&...> and can only be assigned from another tuple<U&...>.

数组的情况下，您可以编写一个函数将该数组变成引用的元组：

With the array case, you can write a function to turn that array into a tuple of references into it:

template <typename T, size_t N, size_t... Is>
auto as_tuple_impl(T (&arr)[N], std::index_sequence<Is...>) {
    return std::forward_as_tuple(arr[Is]...);
}

template <typename T, size_t N>
auto as_tuple(T (&arr)[N]) {
    return as_tuple_impl(arr, std::make_index_sequence<N>{});
}

std::tie(x1, y1, z1) = as_tuple(a); // ok

---

或者，如果您知道有多少个绑定（无论如何都必须要绑定），您可以使用结构化绑定作为返回元组。但是，您必须同时指定大小并为每个大小写一个大小写：

---

Alternatively, if you know how many bindings there are (which you have to anyway), you can use structured bindings as give back a tuple. But you have to both specify the size and write out a case for each one:

template <size_t I, typename T>
auto as_tuple(T&& tuple) {
    if constexpr (I == 1) {
        auto&& [a] = std::forward<T>(tuple);
        return std::forward_as_tuple(a);
    } else if constexpr (I == 2) {
        auto&& [a, b] = std::forward<T>(tuple);
        return std::forward_as_tuple(a, b);
    } else if constexpr (I == 3) {
        // etc.
    }
}

std::tie(x1, y1, z1) = as_tuple<3>(a); // ok

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