迭代器随后解除引用“。”。运算符对。 “ - >”运算符 [英] Iterator de-reference then "." operator versus. "->" operator

问题描述

让我先说这个，我说我通常在c开发，我只熟悉c ++。

我已经得到一些c ++代码，它们具有以下struct的列表/迭代器。

  typedef struct {
 int x; 
 int y; 
} my_struct; 
 
 std :: list< my_struct>列表; 
 std :: list< my_struct> :: iterator the_iter = the_list.begin（）; 
  

然后代码访问 the_iter 这样：

 （* the_iter）.x; 
（* the_iter）.y; 
  

我想将这些更改为更易读的版本：

  the_iter-> x; 
 the_iter-> y; 
  

从我的角度来看，这对于指针解引用是完全正确的。迭代器也是这样吗？是否有任何理由为什么我的同事会使用（*指针）。而不是 p>

考虑到一般情况，可能发生一些迭代器类没有提供 operator - > $ c>，并且执行（* it）.x 将是唯一可能的方法。另一种可能性是 operator * 和 operator - > 有一些非标准语义，不能互换。但是，这个类不能满足任何迭代器的概念，而且技术上不会是迭代器。

在你的情况下，它是 std： ：list< T> :: iterator ，其中 it-> x 和 / code>是等效的。

Let me preface this by saying I usually develop in c and am only familiar with c++.

I've been given some c++ code that has a list/iterator of the following struct.

typedef struct{
  int x;
  int y;
}my_struct;

std::list<my_struct> the_list;
std::list<my_struct>::iterator the_iter = the_list.begin();

The code then accesses x and y of the_iter this way:

(*the_iter).x;
(*the_iter).y;

I want to change these to the more readable version:

the_iter->x;
the_iter->y;

From my c perspective, this is totally fine for pointer dereferencing. Is this also the case for iterators? Is there any reason why my colleague would use (*pointer). instead of p->

解决方案

Considering the general case, it could happen that some iterator class did not provide operator ->, and doing (*it).x would be the only possible way. Another possibility is that operator * and operator -> have some non-standard semantics and are not interchangeble. However, this class would fail to satisfy any iterator concept, and, technically, would not be an iterator.

In your case it is std::list<T>::iterator, for which it->x and (*it).x are equivalent.

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