难道一个类型要求，以宣布它的一个阵列默认构造函数？ [英] Does a type require a default constructor in order to declare an array of it?

问题描述

我注意到，当你声明一个数组，必须所需的默认构造函数。是对的吗？
有没有例外？

I noticed that when you declare an array, the default constructor must be needed. Is that right? Is there any exception?

例如，

struct Foo{
Foo(int i  ) {}
};

int main () {
    Foo f[5];
    return 0;
}

在code以上不能编译。

The code above does not compile.

推荐答案

其他答案都是正确的，但出于完整性：您还可以使用数组初始化语法：

Other answers are all right but, for completeness: You could also use the array initialization syntax:

Foo f[5] = {1,2,3,4,5};

这工作，如果Foo的构造函数不明确。如果是，你必须是明确的....：

This works if Foo's ctor is not explicit. If it was, you'd have to be.... explicit:

Foo f[5] = {Foo(1), Foo(2), Foo(3), Foo(4), Foo(5)};

注意 ¹ ：有两种情况之间的差异，可能不是很明显，因此值得关注：第一，数组元素是直接从构造在 INT ■在初始化列表，通过调用美孚（INT）构造函数。第二，初始化列表是由富第构建与明确美孚（INT）构造函数，和数组元素的复制从初始化列表中的元素构造的。从而为富副本构造函数，需要在后一种情况下

Note¹: There is a difference between the two cases that may not be obvious and is thus worth noting: In the first, the array elements are directly constructed from the ints in the initialization list, by invoking the Foo(int) ctor. In the second, the initialization list is made of Foos constructed with the explicit Foo(int) ctor, and the array elements are copy constructed from the elements in the initialization list. A copy ctor for Foo is thus required in the latter case.

[1]由于MSalters的评论。

[1] Thanks to MSalters for the comment.

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