有一个快捷方式来decltype [英] Is There a Shortcut to decltype

问题描述

在此答案中，我写了C ++ 17代码：

In this answer I wrote the C++17 code:

cout << accumulate(cbegin(numbers), cend(numbers), decay_t<decltype(numbers[0])>{});

这收到一些关于C ++的类型关联的负面评论，我很遗憾地说我同意：（

This received some negative commentary about the nature of C++'s type association, which I'm sad to say that I agree with :(

decay_t< decltype（numbers [0]）> {} 方式获取：

数字的元素的零初始化类型

可以保持与数字'类型的关联，但不能输入30个字符来获取它？

Is it possible to maintain the association with the type of numbers' elements, but not type like 30 characters to get it?

编辑：

包含 accumulate 或从 numbers [0]中提取类型的很多答案。 。问题是他们需要读取器导航到辅助位置以读取解决方案，该解决方案不比初始化代码 decay_t {} 。

I've got a lot of answers involving the a wrapper for either accumulate or for extracting the type from numbers[0]. The problem being they require the reader to navigate to a secondary location to read a solution that is no less complex than the initialization code decay_t<decltype(numbers[0])>{}.

唯一的原因是我们不得不这样做： decltype（numbers [0]）是因为数组下标运算符返回引用：

The only reason that we have to do more than this: decltype(numbers[0]) Is because the array subscript operator returns a reference:

错误：类型为'int'的类型为'int&'的右值表达式的无效转换

error: invalid cast of an rvalue expression of type 'int' to type 'int&'

很有趣的是，对于 decltype 的参数：

It's interesting that with respect to decltype's argument:

如果对象的名称加括号，作为普通的左值表达式

If the name of an object is parenthesized, it is treated as an ordinary lvalue expression

但是， decltype（（numbers [0]））仍然只是对数字的元素的引用。所以最后这些答案可能就像我们可以简化这个初始化一样：（

However, decltype((numbers[0])) is still just a reference to an element of numbers. So in the end these answers may be as close as we can come to simplifying this initialization :(

推荐答案

个人偏好：舞蹈很烦人和困难的 decay_t ， decltype 和 。

Personal preference: I find the decay_t, decltype and declval dance pretty annoying and hard to read.

相反，我会使用一个额外的间接级别通过type-trait value_t< it> 和 init = R {}

Instead, I would use an extra level of indirection through a type-trait value_t<It> and zero-initialization through init = R{}

template<class It>
using value_t = typename std::iterator_traits<It>::value_type;

template<class It, class R = value_t<It>>
auto accumulate(It first, It last, R init = R{}) { /* as before */ }

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