c++ - 用逗号相互依赖初始化？

Question

下面是否完美定义:

int x = 42, y = x;

即严格等价于:

int x = 42;
int y = x;

编辑:问题与风格无关(我知道这是错误的......)，问题是“理论上的”

Answer 1

正确答案是

int x = 42, y = x;

和

int x = 42;
int y = x;

通常是等价的(不严格)。

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考虑到标准 § 8 声明符 [dcl.decl]:

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:

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;

is usually equivalent to

T D1; T D2; ... T Dn;

where T is a decl-specifier-seq and each Di is an init-declarator.

• 以上保证 x = 42 和 y = x 将被单独评估。然而，作为@Praetorian在评论中正确指出，脚注不规范。

• 这意味着评估的顺序没有明确定义，实现者也可以以相反的顺序实现声明的评估(即。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求值。然而，事实并非如此。根据 K & R [K & R II, 3.6 p.63]，这也适用于 C++:

The commas that separate function arguments, variables in declarations, etc., are not comma operators, and do not guarantee left to right evaluation.