覆盖迭代器的运算符*（） [英] Overriding operator*() of iterators

问题描述

假设我有一些用户定义的复杂结构，比如

Say that I have some user-defined complex struct, like

struct usrStruct{
     double a;
     T1 b;
     T2 c;
     /* and so on...*/
}

这是用作std :: vector，std :: list或任何 iterable 的基本元素..

which is used as a basic element for a std::vector, std::list or anything iterable..

假设 std :: vector< usrStruct> 通过迭代器传递给我的函数

Say that the std::vector<usrStruct> is passed to a function of mine through iterators

  template<class InputIterator>  
  T myFoo( InputIterator first, InputIterator last ){ /*...*/ }.

问：是否有标准方法可以覆盖运算符*（）的 InputIterator ，（在这种情况下为 std :: vector< usrStruct> :: iterator ）以便 myFoo 只需与 成员交互 ？

Q: Is there a standard way to override operator*() of the InputIterator, (in this case of std::vector<usrStruct>::iterator ) so that myFoo just interacts with the member a?

即，

  *first == (*first).a;

因此 myFoo 正常工作 关于 usrStruct 的整个定义？

and thus myFoo works orthogonally with respect to the whole definition of usrStruct?

谢谢。

推荐答案

不，你做不到。您可以将结构隐式转换为double（通过运算符double ）。或者你可以通过重载 operator ==（usrSruct，double）和/或 operator ==（double，usrStruct）。

No, you cannot do that. You can make your struct implicitly convertible to a double (via operator double). Or you can allow direct comparisons by overloading operator==(usrSruct,double) and/or operator==(double,usrStruct).

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