C语言 是否需要括号来从指向结构的指针获取结构成员的地址，如“&(s->var)”与“&s->var”？
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Are parenthesis needed to get the address of a struct member from a pointer to the struct as in "&(s->var)" vs "&s->var"?
提问by joker
I have a struct str *s;
我有一个struct str *s;
Let varbe a variable in s. Is &s->varequal to &(s->var)?
让var成为 中的一个变量s。是&s->var等于&(s->var)？
回答by In silico
Behavior-wise, yes they are equivalent since the member access ->operator has a higher precedence than the address-of &operator.
行为方面，是的，它们是等效的，因为成员访问->运算符的优先级高于地址&运算符。
Readibility-wise, the second one &(s->var)is much more readable than &s->varand should be preferred over the first form. With the second form, &(s->var), you won't have to second-guess what it's actually doing as you know the expression in the parentheses are always evaluated first. When in doubt, use parentheses.
在&(s->var)可读性方面，第二种&s->var形式比第一种形式更具可读性，并且应该优先于第一种形式。使用第二种形式 ，&(s->var)您不必再猜测它实际在做什么，因为您知道括号中的表达式总是首先计算。如有疑问，请使用括号。
回答by Smokey.Canoe
Yes.
是的。
(-> is higher precedence than &. See http://cppreference.com/wiki/language/operator_precedence)
（-> 的优先级高于 &。参见http://cppreference.com/wiki/language/operator_precedence）
回答by Christo
Yes, because the pointer dereference operator ->has higher precedence than the address operator &.
是的，因为指针解引用运算符的->优先级高于地址运算符&。
