结构和指针运算阵列 [英] Array of structures and pointer arithmetic

问题描述

是如何使用指针运算访问结构数组？

假设我有一个结构

 结构点{
INT X;
诠释Ÿ;
}收藏[100];
 

假设我有一个函数

  INT FUNC（结构点*集合，INT大小）
 

在这个函数我访问元件，如下所示。

 集合[0] .X
 

这是一样的 *（集合+ 0）.X ？因为。运营商具有较高的precedence比 * 运营商，先收集指针由0增加，和点运算应用，然后将指针取消引用？不知怎的，这是没有意义的;任何澄清是AP preciated。

解决方案

  

这是一样的 *（集合+ 0）.X ？

没有。你的解释是绝对正确的，。具有较高的precedence比 * ，使第二前pression被解析为 *（（收集+ 0）.X）。 集合[I] .X ，另一方面相当于（*（集合+ I））。X 。

其实这尴尬的原因是在 - ＆GT; 运营商推出，所以假设是是一些非琐碎的前pression，你可以写

  Y-X的催化剂
 

而不是

 （*（Y）），X
 

虽然明显集合[0] .X 较干净多了（收集+ 0） - ＆GT; X 在此特定实例

How are arrays of structures accessed using pointer arithmetic?

suppose I have a struct

struct point{
int x;
int y;
}collection[100];

Suppose I have a function

int func(struct point *collection,int size)   

Inside this function I access the element as shown below.

collection[0].x 

Is this the same as *(collection + 0).x? Since the . operator has higher precedence than the * operator, first the collection pointer is incremented by 0, and the the dot operator is applied, and then the pointer dereferenced? Somehow this does not make sense; any clarification is appreciated.

解决方案

Is this the same as *(collection + 0).x?

No. Your explanation is absolutely correct, . has higher precedence than *, so that second expression is parsed as *((collection + 0).x). collection[i].x on the other hand is equivalent to (*(collection + i)).x.

Actually this awkwardness is the reason the -> operator was introduced, so assuming y is some non-trivial expression, you can write

y->x

instead of

(*(y)).x

Although obviously collection[0].x is much cleaner than (collection + 0)->x in this particular instance.

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