没有使用向量迭代器，c ++

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我试图迭代一个向量，它包含指向Student类型对象的指针。 向量的声明如下： static vector< Student *>学生;

I'm trying to iterate over a vector holding a pointer to an object of type Student. The declaration of vector is as follow: static vector<Student*> students;

无论如何，我想在函数pickWinners（）中使用迭代器：

Anyhow, I am trying to use an iterator in function pickWinners():

vector<Student*>::iterator p1 = students.begin(); vector<Student*>::iterator p2 = p1; p2++;

据我所知，p1是指向Student的指针。但是当我尝试这个（例如）：

As I understand, p1 is a pointer to a pointer to Student. But when I try this (for example):

*p1->print();

我得到下一个错误：

Hire.cpp：192：error：请求成员'print'在'* p1 .__ gnu_cxx :: __ normal_iterator :: operator-> with _Iterator = Student **，_Container = std :: vector>'，这是非类类型'Student *' make： * [Hire.o]错误1

Hire.cpp:192: error: request for member ‘print’ in ‘* p1.__gnu_cxx::__normal_iterator<_Iterator, _Container>::operator-> with _Iterator = Student**, _Container = std::vector >’, which is of non-class type ‘Student*’ make: * [Hire.o] Error 1

这对我没有任何意义。我知道问题不是在print（）。 我尝试了

This doesn't make any sense to me. I know the problem is not in print(). I tried

Student *student = students.at(0); student->print();

，一切工作完美。我在这里很无能，任何想法？ 谢谢！

and everything worked perfect. I'm pretty clueless here, any ideas? Thanks!

推荐答案

期望的结果将通过

(*p1)->print();

在这种情况下，代码解析为 *（p1-> print因为 operator-> 的优先级高于 operator * 例如，维基百科上的优先级表

In your case, the code parses as *(p1->print());, because operator-> has higher precedence than operator*, see for example, the precedence table on wikipedia