用逗号相互初始化吗? [英] Interdependent initialization with commas?

问题描述

以下定义是否完美:

int x = 42, y = x;

即严格等同于:

int x = 42;
int y = x;

问题不是关于样式(我知道这是错误的...)，问题是理论上的"

EDIT : the question is not about style (I know that it's wrong...), the question is "theoretical"

推荐答案

正确的答案是

int x = 42, y = x;

和

int x = 42;
int y = x;

通常是等效的(不严格).

考虑标准§8声明符[dcl.decl] :

3分别对声明中的每个init-declarator进行分析，就好像它本身在声明中一样.

3 Each init-declarator in a declaration is analyzed separately as if it was in a declaration by itself.

并在脚注[100]中进一步说明:

and in the footnote [100] further explains:

具有多个声明符的声明通常等效于相应的声明序列，每个声明都有一个 声明者.那是

A declaration with several declarators is usually equivalent to the corresponding sequence of declarations each with a single declarator. That is

T D1，D2，... Dn;

通常等同于

T D1; T D2； ... T Dn;

，其中T是decl-specifer-seq，每个Di是init声明符.

• 以上保证了x = 42和y = x将被分别求值.但是，正如 @Praetorian 在评论中正确指出的那样，脚注不是规范性的.

  • The above guarantees that x = 42 and y = x will be evaluated separately. However, as @Praetorian correctly pointed out in the comments, footnotes are not normative.

    这意味着评估的顺序没有很好地定义，实现者也可以以相反的顺序(即T Dn; ...T D2; T D1;)实现声明的评估.

    This means that the order of evaluation is not well defined and an implementer could as well implement the evaluation of the declarations in the reverse order (i.e,. T Dn; ...T D2; T D1;).

    有人可能会争辩说逗号运算符可以保证从左到右的评估.但是，事实并非如此.根据K& R [K& R II，第3.6页，第63页]，也适用于C ++:

    One might argue that the comma operator is guaranteed left to right evaluation. However, this not the case. According to the K & R [K & R II, 3.6 p.63], that also applies to C++:

    分隔函数参数，声明中的变量等的逗号不是逗号运算符，也不保证从左到右的求值.

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