C++ 重载运算符+;:表达式必须具有整型或非作用域枚举类型
我正在编写一些使用链表的代码,我很难确定为什么我的+运算符不起作用,我一直收到上面的错误 我错过了什么C++ 重载运算符+;:表达式必须具有整型或非作用域枚举类型,c++,linked-list,operator-overloading,C++,Linked List,Operator Overloading,我正在编写一些使用链表的代码,我很难确定为什么我的+运算符不起作用,我一直收到上面的错误 我错过了什么 #include <iostream> #include <string> using namespace std; struct sortedListNode { char letter; int occurrences = 1; sortedListNode *next; }; sortedListNode *operator+(sort
#include <iostream>
#include <string>
using namespace std;
struct sortedListNode
{
char letter;
int occurrences = 1;
sortedListNode *next;
};
sortedListNode *operator+(sortedListNode *lhs, sortedListNode *rhs)
{
int i = 0;
sortedListNode *head, *tail, *curr, *prev, *tempLoc;
sortedListNode *list1, *list1curr, *list2, *list2curr;
list1 = lhs;
list2 = rhs;
// Copy list1 into output list
head = new sortedListNode;
head->letter = list1->letter;
head->occurrences = list1->occurrences;
head->next = NULL;
tail = head;
list1curr = list1;
list1curr = list1curr->next;
while (list1curr != NULL)
{
tempLoc = new sortedListNode;
tempLoc->letter = list1curr->letter;
tempLoc->occurrences = list1curr->occurrences;
tempLoc->next = NULL;
tail->next = tempLoc;
tail = tempLoc;
list1curr = list1curr->next;
}
curr = head;
while (list2curr != NULL)
{
while (curr != NULL)
{
if (curr->letter == list2curr->letter)
{
curr->occurrences++;
break;
}
else if ((curr->letter > list2curr->letter) && (curr == head))
{
tempLoc = new sortedListNode;
tempLoc->next = curr;
tempLoc->letter = list2curr->letter;
head = tempLoc;
break;
}
else if ((curr->letter > list2curr->letter) && (curr != head))
{
tempLoc = new sortedListNode;
tempLoc->next = curr;
tempLoc->letter = list2curr->letter;
prev->next = tempLoc;
break;
}
else if ((curr == tail) && (curr->letter < list2curr->letter))
{
tempLoc = new sortedListNode;
tempLoc->next = NULL;
tempLoc->letter = list2curr->letter;
tail->next = tempLoc;
tail = tempLoc;
break;
}
prev = curr;
curr = curr->next;
}
curr = head;
list2curr = list2curr->next;
}
return head;
}
sortedListNode *fromString(string inWord)
{
int i = 0;
sortedListNode *head, *tail, *curr, *prev, *tempLoc;
// Put the first letter in as the first element, set head and
// tail to this element.
head = new sortedListNode;
tail = head;
head->letter = inWord[0];
head->next = NULL;
curr = head;
for (int i = 1; inWord[i] != '\0'; i++)
{
while (curr != NULL)
{
if (curr->letter == inWord[i])
{
curr->occurrences++;
break;
}
else if ((curr->letter > inWord[i]) && (curr == head))
{
tempLoc = new sortedListNode;
tempLoc->next = curr;
tempLoc->letter = inWord[i];
head = tempLoc;
break;
}
else if ((curr->letter > inWord[i]) && (curr != head))
{
tempLoc = new sortedListNode;
tempLoc->next = curr;
tempLoc->letter = inWord[i];
prev->next = tempLoc;
break;
}
else if ((curr == tail) && (curr->letter < inWord[i]))
{
tempLoc = new sortedListNode;
tempLoc->next = NULL;
tempLoc->letter = inWord[i];
tail->next = tempLoc;
tail = tempLoc;
break;
}
prev = curr;
curr = curr->next;
}
curr = head;
}
return head;
}
void printList(sortedListNode *inSortedListNode)
{
sortedListNode *curr;
curr = inSortedListNode;
int nodeCounter = 0;
while (curr != NULL)
{
nodeCounter++;
cout << "Node " << nodeCounter << " at " << curr << " - Letter = " << curr->letter << ", Occurrences = " << curr->occurrences << ", Next Node = "
<< curr->next << endl;
curr = curr->next;
}
}
int main()
{
string word1, word2;
sortedListNode *list1;
sortedListNode *list2;
sortedListNode *list3;
cout << "Enter first word: ";
cin >> word1;
cout << "Enter second word: ";
cin >> word2;
list1 = fromString(word1);
cout << "Letter list from word one: " << endl;
printList(list1);
list2 = fromString(word2);
cout << "Letter list from word two: " << endl;
printList(list2);
list3 = list1 + list2;
cout << "Letter list from both words: " << endl;
printList(list3);
cin.ignore(32767, '\n');
char dummy[1]{};
cin.getline(dummy, 1);
}
#包括
#包括
使用名称空间std;
结构分类列表节点
{
字符字母;
int=1;
sortedListNode*下一步;
};
sortedListNode*操作员+(sortedListNode*lhs、sortedListNode*rhs)
{
int i=0;
分类列表节点*头、*尾、*当前、*上一个、*模板;
sortedListNode*list1、*list1curr、*list2、*list2curr;
列表1=lhs;
列表2=rhs;
//将列表1复制到输出列表中
head=新的分拣列表节点;
标题->字母=列表1->字母;
头部->出现次数=列表1->出现次数;
head->next=NULL;
尾=头;
list1curr=list1;
list1curr=list1curr->next;
while(list1curr!=NULL)
{
tempLoc=新的sortedListNode;
tempLoc->letter=list1curr->letter;
tempLoc->executions=list1curr->executions;
tempLoc->next=NULL;
tail->next=tempLoc;
tail=tempLoc;
list1curr=list1curr->next;
}
curr=头;
while(list2curr!=NULL)
{
while(curr!=NULL)
{
if(curr->letter==list2curr->letter)
{
curr->occurrents++;
打破
}
如果((货币->字母->列表2货币->字母)&&(货币==标题))
{
tempLoc=新的sortedListNode;
tempLoc->next=curr;
tempLoc->letter=list2curr->letter;
头部=tempLoc;
打破
}
else if((curr->letter>list2curr->letter)&(curr!=head))
{
tempLoc=新的sortedListNode;
tempLoc->next=curr;
tempLoc->letter=list2curr->letter;
上一步->下一步=tempLoc;
打破
}
如果((curr==tail)和&(curr->letterletter))
{
tempLoc=新的sortedListNode;
tempLoc->next=NULL;
tempLoc->letter=list2curr->letter;
tail->next=tempLoc;
tail=tempLoc;
打破
}
上一次=当前;
当前=当前->下一步;
}
curr=头;
list2curr=list2curr->next;
}
回流头;
}
sortedListNode*fromString(字内字符串)
{
int i=0;
分类列表节点*头、*尾、*当前、*上一个、*模板;
//将第一个字母作为第一个元素,设置开头和结尾
//这个元素的尾部。
head=新的分拣列表节点;
尾=头;
标题->字母=inWord[0];
head->next=NULL;
curr=头;
for(inti=1;inWord[i]!='\0';i++)
{
while(curr!=NULL)
{
if(curr->letter==inWord[i])
{
curr->occurrents++;
打破
}
else if((curr->letter>inWord[i])&&(curr==head))
{
tempLoc=新的sortedListNode;
tempLoc->next=curr;
tempLoc->letter=inWord[i];
头部=tempLoc;
打破
}
else if((curr->letter>inWord[i])&&(curr!=head))
{
tempLoc=新的sortedListNode;
tempLoc->next=curr;
tempLoc->letter=inWord[i];
上一步->下一步=tempLoc;
打破
}
如果((curr==tail)和(&(curr->letternext=NULL;
tempLoc->letter=inWord[i];
tail->next=tempLoc;
tail=tempLoc;
打破
}
上一次=当前;
当前=当前->下一步;
}
curr=头;
}
回流头;
}
无效打印列表(sortedListNode*inSortedListNode)
{
sortedListNode*curr;
curr=inSortedListNode;
int nodeCounter=0;
while(curr!=NULL)
{
nodeCounter++;
不能
你需要
sortedListNode *operator+(sortedListNode*lhs, sortedListNode*rhs);
但你只提供
sortedListNode *operator+(sortedListNode lhs, sortedListNode rhs);
我还应该说,为指针类型重载+
运算符没有任何意义(即使允许).错误消息的直接原因是未调用您定义的运算符+
。这是因为您已将运算符+
定义为将两个sortedListNode
作为参数,但您传递了两个指针。如果您将运算符改为使用指针(sortedListNode*
),它将跳过即时语法错误
然而,这才是真正问题的开始。operator+的实现也有问题。它返回head
,它不是sortedListNode
结构的一部分。至关重要的是,它没有使用传递给它的两个列表中的任何一个
编辑:现在我们可以看到完整的代码,谢谢你发布
我想附和Walter所说的:添加指针毫无意义。相反,您应该定义一个运算符+来将列表添加到一起:
sortedNodeList operator+(const sortedNodeList& lhs, const sortedNodeList &rhs){
...
}
换句话说,您的参数应该是对列表的引用(而不是指针)您的结果应该是一个列表,而不是指针。我很难确定您试图编译的代码。请发布一个。澄清操作符调用在main中,这就是我得到语法错误的原因。+
操作符是否应该将这两个列表关联起来(即将它们链接在一起)或者返回一个与输入列表的concatation相等的新列表?您的代码似乎执行后一种操作,但是参数(lhs
,rhs
)可能是const sortedListNode*
。很抱歉,它应该返回一个新列表,这是正确的。a
sortedNodeList operator+(const sortedNodeList& lhs, const sortedNodeList &rhs){
...
}