是直接初始化禁止数组？ [英] Is direct initialization forbidden for array?

问题描述

语言律师的角度来看，这些条款中禁止低于code标准：

Language-lawyer-wise, which clause in the standard forbid below code:

int arr[] (10, 42); 

这将产生10个元素的数组，每个initalized 42。

This would produce an array of 10 elements, each initalized to 42.

推荐答案

语言律师明智的，8.5 / 17：

Language-lawyer wise, 8.5/17:

- 如果初始值是（非括号的）支撑，初始化列表，对象或引用列表初始化（8.5.4）

— If the initializer is a (non-parenthesized) braced-init-list, the object or reference is list-initialized (8.5.4).

- 如果目标类型是引用类型，见8.5.3

— If the destination type is a reference type, see 8.5.3.

- 如果目标类型是字符数组，char16_t的阵列，char32_t的阵列，或
  wchar_t的阵列，并且初始化为一个字符串，见8.5.2。

— If the destination type is an array of characters, an array of char16_t, an array of char32_t, or an array of wchar_t, and the initializer is a string literal, see 8.5.2.

- 如果初始化为（），对象是值初始化

— If the initializer is (), the object is value-initialized.

- 否则，如果目标类型是一个数组，是形成不良的程序

— Otherwise, if the destination type is an array, the program is ill-formed

一个支撑-初始化列表为{}其中任何（或没有）可以在括号内（例如 INT ARR [3] = {1,2,3} ）。考虑到这一点，没有第4个选项是可行的 INT ARR []（10，42）; ，留下最后一个指示节目是非法的构造。

A braced-init-list is { } where anything (or nothing) can be inside the brackets (for example, int arr[3] = {1,2,3}). With that in mind, none of the first 4 options are viable for int arr[] (10, 42);, leaving the last one indicating the program is ill-formed.

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