[英]LinkedList C++ implementation
我剛剛創建了LinkedList的實現(僅用於自我教育目的。)。 我讓它運行,但輸出結果有點奇怪...這是代碼:
#include "stdafx.h"
#include <iostream>
#include <stdio.h>
using namespace std;
template <class T>
class Node{
T datum;
Node<T> *_next;
public:
Node(T datum)
{
this->datum = datum;
_next = NULL;
}
void setNext(Node* next)
{
_next = next;
}
Node* getNext()
{
return _next;
}
T getDatum()
{
return datum;
}
};
template <class T>
class LinkedList{
Node<T> *node;
Node<T> *currPtr;
Node<T> *next_pointer;
int size;
public:
LinkedList(T datum)
{
node = new Node<T>(datum);
currPtr = node; //assignment between two pointers.
next_pointer = node;
size = 1;
}
LinkedList* add(T datum) // return pointer type.
{
Node<T> *temp = new Node<T>(datum);
currPtr->setNext(temp);
currPtr = temp;
size++;
cout<<datum<<" is added.";
return this; //pointer type specification
}
T next()
{
T data = (*next_pointer).getDatum();
cout<<data<<" is visited.";
next_pointer = next_pointer->getNext();
return data;
}
int getSize()
{
return size;
}
};
現在我嘗試使用LinkedList:
int main()
{
LinkedList<int> *list = new LinkedList<int>(1);
list->add(2)->add(3)->add(4);
cout<<endl;
printf("%d %d %d %d",list->next(),list->next(),list->next(),list->next()); \\One
cout<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<endl; \\Two
cout<<list->next()<<endl;\\Three
cout<<list->next()<<endl;
cout<<list->next()<<endl;
cout<<list->next()<<endl;
}
輸出One將顯示數據:4 3 2 1.兩個將顯示4 3 2 1.三個將顯示1 2 3 4.我不知道運行期間發生了什么。 所有這些都應該以1 2 3 4順序輸出數據...我很感激你的幫助! 謝謝!
未指定參數的評估順序,因此:
printf("%d %d %d %d",list->next(),list->next(),list->next(),list->next());
可以先評估最后一個list->next()
,或者中間的...
編輯:只是解決我認為的問題,因為我懷疑這是實際的代碼: http : //ideone.com/avEv7
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