initializer_list 返回的生命周期延长 [英] Lifetime Extension of a initializer_list return

问题描述

所以我有一个返回类型为 auto 的 lambda，我遇到了支持 initializer_list 的数组在此处被销毁的问题:

So I have a lambda who's return type is auto and I'm having issues with the array backing for an initializer_list being destroyed here:

const auto foo = [](const auto& a, const auto& b, const auto& c) { return {a, b, c}; };

我将像这样使用 lambda:

I will be using the lambda like this:

auto bar = foo(1, 2, 3);

for(const auto& i : bar) cout << i << endl;

我正在从事的一项工作将所有 lambda 表达式作为单个语句作为其编码标准的一部分(请随意表达您的愤怒.)我想我可以通过以下方式解决这个问题:

A job that I'm working has as part of their coding standard that all lambdas are single statements (feel free to express your outrage.) I think that I can get around this by:

1. 给 foo 一个 vector int 的返回类型，但这弄乱了它的通用性:const auto foo = [](const auto& a, const auto& b, const auto& c) ->向量{返回{a，b，c}；}
2. 只需编写一个模板化函数，该函数构造一个 vector 并返回它:template 向量 Tfoo(const T& a, const T& b, const T& c){ return {a, b, c};}

1. Giving foo a return type of vector int, but that messes up how nicely generic it is: const auto foo = [](const auto& a, const auto& b, const auto& c) -> vector<int> { return {a, b, c}; }
2. Just writing a templatized function which constructs a vector<T> and returns it: template <typename T> vector<T> foo(const T& a, const T& b, const T& c){ return {a, b, c}; }

是否可以将这些变量强制放入一个容器中，谁的支持不会在一行中全部销毁，以便我可以保留带有 auto 返回类型的 lambda?

Is it possible to coerce these variables into a container, who's backing won't be destroyed all in one line so that I can keep the lambda with an auto return type?

推荐答案

库基础 TS v2 有 std::experimental::make_array，肯定能满足你的要求:

The library fundamentals TS v2 has std::experimental::make_array, which would certainly satisfy your requirements:

#include <experimental/array>
const auto foo = [](const auto& a, const auto& b, const auto& c) {
    return std::experimental::make_array(a, b, c); };

更一般地，构造函数的模板参数推导 允许你写:

More generally, template argument deduction for constructors would allow you to write:

const auto foo = [](const auto& a, const auto& b, const auto& c) {
    return std::vector{a, b, c}; };
                      ^-- no template parameter required

今天，您可以使用 common_type:

Today, you could emulate this using common_type:

const auto foo = [](const auto& a, const auto& b, const auto& c) {
    return std::vector<std::common_type_t<decltype(a), decltype(b), decltype(c)>>{a, b, c}; };

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