为什么他们特殊的初始化列表而不是一样的？ [英] Why would they special-case certain initializer lists instead of treating them all the same?

问题描述

假设我有一个变量 auto x ，我想使用大括号初始化初始化为 7 / p>

Say I have a variable auto x that I want to initialize to 7 using brace initialization, simple:

auto x {7};

除非我知道x是 NOT 一个整数，本身。为什么？是否有一个具体原因，为什么委员会会决定 auto 应该在单个自动值的情况下抓取初始化列表，或者他们希望我们只是实现这些 t一起使用。 我似乎没有想到一个可能的原因，我想要一个初始化列表存储到自动而不是值

Except I learned that x is NOT an integer, but an initialization list itself. Why? Is there a specific reason why the committee would decide that auto should grab the initialization list in the case of a single auto value, or do they expect us to just realize these shouldn't be used together. I cant seem to think of a possible reason i would want an initializer list to be stored into auto as opposed to the value

推荐答案

一个非常实用的答案是，为什么应该选择 int ？或 double ，或任何带有int单参数构造函数的UDT？通过拒绝推导一个具体类型，编译器保留了更一般的初始化列表的任何可能的应用。

A very practical answer is, "why should int be selected?" Or double, or any UDT with an int single-argument constructor? By declining to deduce a concrete type, the compiler preserves any possible application of the more general initializer list.

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