在括号需要从一个指针得到一个结构成员的地址结构体如在QUOT;及（S-GT＆; VAR）QUOT; VS＆QUOT;＆安培; S-＆GT; VAR＆QUOT;？ [英] Are parenthesis needed to get the address of a struct member from a pointer to the struct as in "&(s->var)" vs "&s->var"?

问题描述

我有一个结构STR * S ;

让 VAR 在 A变量s 。为＆放大器; S-＆GT; VAR 等于及（S-＆GT; VAR）？

推荐答案

行为的角度来看，是的，他们是因为等价的成员访问 - ＆GT; 经营者具有更高的precedence比地址的＆安培; 运营商。

Behavior-wise, yes they are equivalent since the member access -> operator has a higher precedence than the address-of & operator.

可读性明智的，第二个及（S-＆GT; VAR）比＆放大器更可读; S-＆GT; VAR 并应pferred在第一种形式$ p $。对于第二种形式，及（S-＆GT; VAR），你就不必去猜测它实际上是在做什么，你知道前pression括号总是首先评估。如有疑问，请使用括号。

Readibility-wise, the second one &(s->var) is much more readable than &s->var and 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.

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