当我在编译时不知道时，如何从 std::tuple 获取第 i 个元素?

2022-01-20 00:00:00 tuples c++ c++11

我有一个 std::size_t 类型的变量 i 和一个 std::tuple 类型的元组.我想获取元组的 i-th 元素.我试过这个:

I have a variable i of type std::size_t and a tuple of type std::tuple. I want to get the i-th element of the tuple. I tried this:

// bindings... is of type const T&... auto bindings_tuple = std::make_tuple(bindings...); auto binding = std::tuple_element<i, const T&...>(bindings_tuple);

但是我得到这个编译错误，说第一个模板参数必须是一个整数常量表达式:

But I get this compile error saying that the first template argument must be an integral constant expression:

错误:'std::size_t' 类型的非类型模板参数(又名'unsigned long')不是整数常量表达式

error: non-type template argument of type 'std::size_t' (aka 'unsigned long') is not an integral constant expression

是否可以获取元组的第 i 元素，以及如何获取?

Is it possible to get the i-th element of a tuple, and how to do that?

_{^{如果可能的话，我想不使用 boost.}}

推荐答案

你不能.这不是元组的用途.如果你需要动态访问一个元素，使用 std::array<T,N>，它几乎等同于 std::tuplecode> 但为您提供动态 [i]-运算符；甚至是像 std::vector<T> 这样的全动态容器.
You cannot. That's not what a tuple is for. If you need dynamic access to an element, use std::array<T,N>, which is almost identical to std::tuple<T,...,T> but gives you the dynamic [i]-operator; or even a fully dynamic container like std::vector<T>.

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