用于向量遍历的C ++迭代器，可能的问题 [英] C++ iterator for vector traversal, possible issue

问题描述

大家好！



所以我还有另一个问题。



我必须遍历矢量并使用迭代器打印出这些值。



我知道你说....这有多难？



我的想法一样，所以我喜欢这个

Hi all!

So I have another problem.

I have to traverse the vector and print out the values using an iterator.

I know you say.... how hard can that be?

I thought the same so i did like this

// suppose we have a vector that takes a pointer to an object of abstract class

// like this

std::vector<foo*> myVector(5); // declared a vector

// a print method just prints data of each object in the vector

for(int a = 0; a < myVector.size()-1; a++){
   myVector.at(a)->print();
}

// this works but using an iterator doesnt. Why?

// suppose we have this traversal

for(std::vector<foo*>::iterator a = myVector.begin(); i != myVector.end(); ++a){
 // trying to print out each object

  cout << *a << endl;

// of course it won't work why? first can't even get to print function

// second trying to access the object of an abstract class by derefencing the pointer.

// Oh and this, by dereferencing i cout the address of it but can't print out with print function
}

好​​的，我希望我没有犯过任何粗心的错误。



请检查我的代码并告诉我我是否正在做任何事情错误。



据我所知，我不能取消引用抽象类的指针，因为这样会留下一个对象，但是这个对象不存在。



用iterator可以做到吗？



提前致谢！



我尝试了什么：



尝试了解问题的方式。

Ok i hope i didn't make any careless mistakes.

Please check my code and tell me whether i'm doing anything wrong.

As far as i know i can't dereference a pointer to abstract class since that would leave me with an object but this object doesn't exsist.

Is it possible to do it with iterator?

Thanks in advance!

What I have tried:

Tried the way it is written in the body of the problem.

推荐答案

for(int a = 0; a < myVector.size()-1; a++){
   myVector.at(a)->print();
}

你确定这个循环不会错过向量中的最后一个位置吗？

[UpDate]

Are you sure this loop don't miss the last place in the vector ?
[UpDate]

Quote：

它从零到大小 - 1

It goes from zero to size - 1

你的代码停止在 size-1 之前，它停在 size-2 ，再次检查。

your code stops before size-1, so it stops at size-2, check again.

您可能想要

cout << (*a)->print() << endl;

实际上 * a 将是指向<$ c的指针$ c> foo 。



就指针指向某个东西，你可以取消引用它。关于抽象类指针的注释没有任何意义。

In fact *a would be a pointer to foo.

As far a pointer point to something, you can dereference it. The comment about a pointer to abstract class does not make any sense.

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