具有运算符重载功能的c ++ 17评估顺序 [英] c++17 evaluation order with operator overloading functions

问题描述

关于这个问题

C ++引入了哪些评估顺序保证17?

具有此规范

http://www.open- std.org/jtc1/sc22/wg21/docs/papers/2016/p0145r3.pdf

这是规范中的文字

此外，我们建议以下附加规则: 包含重载运算符的表达式的求值是 由与相应内置组件关联的顺序确定 运算符，而不是函数调用规则.

Furthermore, we suggest the following additional rule: the order of evaluation of an expression involving an overloaded operator is determined by the order associated with the corresponding built-in operator, not the rules for function calls.

这是否意味着这两个表达式不再等效?

Does this mean that these two expressions are no longer equivalent?

a << b;
operator<<(a, b);

由于第二个函数看起来像函数调用，因此参数中没有确定的评估顺序?

As the second one looks like a function call, hence there is no guaranteed evaluation order in the parameters?

推荐答案

由于第二个函数看起来像函数调用，因此参数中没有确定的评估顺序?"

的确. [expr.call]/5 包含一个示例，专门介绍了两者之间的区别您的问题[强调我的]中涉及的案例:

Indeed. [expr.call]/5 contains an example specifically covering the difference between the two cases covered in your question [emphasis mine]:

postfix-expression在 表达式列表和任何默认参数.初始化 参数，包括每个关联的值计算和边 效果，相对于其他任何元素不确定地排序 参数.

The postfix-expression is sequenced before each expression in the expression-list and any default argument. The initialization of a parameter, including every associated value computation and side effect, is indeterminately sequenced with respect to that of any other parameter.

...

注意:如果使用运算符符号调用了运算符功能， 参数评估按内置指令的指定顺序进行 运营商；看 [over.match.oper] . [示例:

Note: If an operator function is invoked using operator notation, argument evaluation is sequenced as specified for the built-in operator; see [over.match.oper]. [ Example:

struct S {
  S(int);
};
int operator<<(S, int);
int i, j;
int x = S(i=1) << (i=2);
int y = operator<<(S(j=1), j=2);

执行初始化后，i的值为2(请参见 [expr.shift] )，但未指定 > j的值是否为 1或2.

After performing the initializations, the value of i is 2 (see [expr.shift]), but it is unspecified whether the value of j is 1 or 2.

-示例]

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