问题描述
标准提到 f(a，(t = 3，t + 2)，c); 根据我的理解，这将是赋值表达式，后跟第二个运算符的表达式.
The standard mentions f(a,(t=3,t+2),c); which would be an assignment-expression followed by an expression for the 2nd operator according to my understanding.
但是语法将它并列列出:
But the grammar lists it juxtaposed:
表达式:
expression:
作业表达
assignment-expression
表达式，赋值表达式
expression, assignment-expression
工作草案，编程标准语言C++修订版N4140(2014年11月)
Working Draft, Standard for ProgrammingLanguage C++ Revision N4140 (November 2014)
有人很高兴向我解释我在这里想念的是什么吗?
Is someone so nice as to explain to me please what it is that I'm missing here?
推荐答案
看到时
expression:
assignment-expression
expression, assignment-expression
这意味着有两种表达方式.一种可能是 assignment-expression ，它是在较早的位置定义的.或者将其递归表示为 expression, assignment-expression
It mean that there are 2 possibilities for expression. One possibility that it is just assignment-expression that is defined somewhere earlier. Or it is recursively represented as expression, assignment-expression
因此，将其扩展后，您会收到一个逗号分隔的一个或多个赋值表达标记的列表.
So after extending it you receive that expression is comma separated list of one or more assignment-expression tokens.
在示例中，您提到的第二个参数是表达式(t = 3，t + 2)，它由2个逗号分隔的赋值表达式组成-由于它显示为在给逗号赋予特殊含义的上下文中"，必须仅出现在括号中".
In the sample you're mentioned second parameter is expression (t=3,t+2) which consists of 2 comma-separated assignment-expressions - and since it appears "In contexts where comma is given a special meaning" it has to "appear only in parentheses".
要弄清赋值表达式为什么可以采用t + 2的形式，您必须从其定义中返回并始终选择首选项
To find out why assignment-expression could take a form of t+2 you have to go back from its definitions and choose first choice always
assignment-expression
-> conditional-expression
--> logical-or-expression
---> logical-and-expression
----> inclusive-or-expression
-----> exclusive-or-expression
------> and-expression
-------> equality-expression
--------> relational-expression
---------> shift-expression
----------> additive-expression - this is what you see
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