直接初始化和复制引用的初始化 [英] direct initialization and copy initialization of reference

问题描述

我一直在读C ++引物。对于对象的初始化，C ++支持两种形式的初始化：直接和复制。
但是书中没有引用初始化的引用。在书中我从来没有见过直接初始化（如果存在）的引用。所有是复制一个像：

I have been reading "C++ primer". For the initialization of object, C++ supports 2 forms of initialization: direct and copy. but the book does not refer the initialization of reference. And in the book I have never seen the direct initialize(if exists) of a reference. All is the copy one like:

int i;
int &j = i;//but not int &j(i);which also works in my experiment

我想知道的是，它是相同的下面为初始化一个引用。
代表以下代码：

I want to know that is it the same that are going on underneath for the initialization of a reference. for the following codes:

string null_book = "9-999-99999-9";

初始化进度首先创建一个临时字符串对象tmp（例如），将直接初始化为ac样式字符串参数，然后使用副本构造函数初始化变量null_book。这对我有意义。
对于这一个：

the initialization progress is first create a temporary string object tmp(for instance) that will direct initialized with a c style string parameter, and then initialize the variable null_book with the copy Constructor. That make sense to me. for this one:

int &j = i;

将以同样的方式初始化ref j？这将是一个由i初始化的温度参考（例如），然后用t初始化j。这不是有意义吗？为什么书从不使用直接初始化来参考？
感谢您的关注！

will ref j be initialized the same way? That will be a temp reference it &t(for example) initialized by i and then initialize j with t? that doesnt make sense??? Why the book never use the direct initialization for reference? Thanks for your attention!

推荐答案

直接初始化和复制初始化的主要区别在8.5节，标准第17段。一般来说，区别在于对于复制初始化中的类类型，不考虑显式构造函数（仅考虑转换构造函数），并创建可能的复制副本;在直接初始化显式构造函数中，并且直接构造目标。从标准的第8.5节：

The main differences between direct-initialization and copy-initialization are covered in section 8.5, paragraph 17 of the standard. In general, the difference is that for class types in copy-initialization, explicit constructors are not considered (only converting constructors are considered) and a possibly elided copy is made; in direct-initialization explicit constructors are considered and the target is constructed directly. From section 8.5 of the standard:

14 - 初始化的形式（使用括号或 = ）通常是无关紧要的，但是当初始化器或被初始化的实体有类类型[...]时非常重要。

14 - The form of initialization (using parentheses or =) is generally insignificant, but does matter when the initializer or the entity being initialized has a class type [...]

对于非类类型（包括引用），直接初始化和复制初始化具有类似的语义;对于引用，引用绑定在两种情况下都会发生，如 8.5.3引用[dcl.init.ref] 中所指定。引用的直接初始化和复制初始化只有涉及转换函数的不同语义（ 13.3.1.6通过直接引用绑定的转换函数初始化[over.match.ref] ）;

For non-class types (including references), direct-initialization and copy-initialization have similar semantics; for references, a reference binding occurs in either case, as specified in 8.5.3 References [dcl.init.ref]. Direct-initialization and copy-initialization of a reference only have different semantics where a conversion function is involved (13.3.1.6 Initialization by conversion function for direct reference binding [over.match.ref]); again, direct-initialization is allowed to invoke explicit conversion functions where copy-initialization is not.

因此，在

int &j = i;

8.5.3p5适用，参考 j 直接绑定到 i 。

8.5.3p5 applies and the reference j is bound directly to the lvalue i. No temporaries are invoked.

在复杂性方面，引用更接近基本类型（原始类型）而不是类类型。基元被初始化而没有临时构造（8.5p17，最后一个项目符号），一般来说引用也是。这可能是为什么本书只使用 = 形式来初始化引用;与原语一样，通常没有区别，并且 int i = x; 通常比 int i（x）; 。

In terms of complexity, references are closer to fundamental (primitive) types than to class types. Primitives are initialized without a temporary being constructed (8.5p17, last bullet) and in general references are too. This is probably why the book only uses the = form for initialization of references; as with primitives, there is usually no difference and writing int i = x; is usually clearer than int i(x);.

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