是直接初始化还是复制初始化？ [英] Is it direct-initialization or copy-initialization?

问题描述

可以使用各种方法来初始化C ++中的对象（类或结构的实例）。某些语法引起对象的直接初始化，其他语法导致复制初始化。在编译器中启用 copy-elision 时，两者的性能相同。禁用 copy-elision 时，在选择每个实例化（复制初始化）时，每个实例化都会有一个附加的copy / move构造函数调用。

Initializing objects (instances of classes or structs) in C++ can be done in various ways. Some syntaxes evoke a direct-initialization of your object, other syntaxes lead to a copy-initialization. With copy-elision enabled in the compiler, both have identical performance. With copy-elision disabled, there is an additional copy/move constructor call upon every instantiation when you choose for the latter (copy-initialization).

结论：复制初始化可能会影响性能！

Conclusion: copy-initialization can have a performance-penalty!

来自以下问题： C ++ 11成员初始值设定项列表与类初始值设定项？我可以得出结论，这将是复制初始化语法：

From the following question: C++11 member initializer list vs in-class initializer? I can conclude that this would be copy-initialization syntax:

obj s = obj("value");

这将是直接初始化语法：

obj s{"value"};

但是这个呢：

 
But what about this one:

obj s = {"value"};

而这个：

obj s = obj{"value"};

而这个：

obj s("value");

或者这一个：

obj s = "value";

---

注意

Bjarne Stroustrup在他的书使用C ++进行编程，原理和实践第二版，第311页，第9.4.2节中比较了几种初始化样式（但不是全部）：

---

NOTE
Bjarne Stroustrup compares a few initialization styles (but not all) in his book "Programming, Principles and Practice Using C++" 2nd edition, on page 311, §9.4.2:

struct Date {
    int y,m,d;                     //year, month, day
    Date(int y, int m, int d);     //check for valid date and initialize
    void add_day(int n);           //increase the Date by n days
};

...

Date my_birthday;                   //error: my_birthday not initialized
Date today{12,24,2007};             //oops! run-time error
Date last{2000,12,31};              //OK (colloquial style)
Date next = {2014,2,14};            //also OK (slightly verbose)
Date christmas = Date{1976,12,24};  //also OK (verbose style)

先生。 Stroustrup将这些不同的初始化样式呈现为相等。至少，这就是我的样子。不过，由于本书中尚未讨论这些术语，因此有些仍然可能是直接初始化，而另一些是 copy-initialization 。

Mr. Stroustrup presents these different initialization styles as equal. At least, that's how it looks to me. Nevertheless, it could still be possible that some are direct-initialization and others copy-initialization, since those terms are not yet discussed at that point in the book.

编辑

给定的答案会带来一些有趣的结果。

显然，这是直接初始化：

obj s("value");

这是直接列表初始化：

obj s{"value"};

有些人指出，这是有区别的。它们实际上有什么不同？

As some of you point out, there is a difference. In what way do they actually differ? Would the difference be noticeable in the output of a non-optimizing compiler?

推荐答案

obj s = obj("value");

这是对prvalue的直接初始化，然后将其用于复制初始化变量 s 。 C ++ 17的prvalue规则实际上对 s 进行了直接初始化。

This is direct initialization of a prvalue, which is then used to copy initialize the variable s. C++17's prvalue rules make this de-facto direct initialization of s.

obj s{"value"};

这是直接- 列表 -初始化。 列表部分很重要。任何时候为了初始化对象而应用了支撑初始化列表，都在执行列表初始化。

This is direct-list-initialization. The "list" part is important. Anytime you apply a braced-init-list for the purposes of initializing an object, you are performing list-initialization.

obj s = {"value"};

这是复制列表初始化。

obj s = obj{"value"};

这是对prvalue的直接列表初始化，即然后用于复制初始化变量 s 。

This is direct-list-initialization of a prvalue, which is then used to copy initialize the variable s.

obj s("value");

这是直接初始化。

obj s = "value";

这是副本初始化。

先生。 Stroustrup会以相同的方式呈现这些不同的初始化样式。

Mr. Stroustrup presents these different initialization styles as equal.

就它们在大多数情况下所做的相同而言，它们是相等的。但是他们在技术上并不平等；复制列表初始化不能调用 explicit 构造函数。因此，如果选择的构造函数是 explicit ，则在复制列表初始化的情况下代码将无法编译。

They are equal in the sense that they do mostly the same thing. But they're not technically equal; copy-list-initialization cannot call explicit constructors. So if the selected constructor were explicit, the code would fail to compile in the copy-list-initialization cases.

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