时间:2019-05-06 标签:c++error: expected type-specifier

当我尝试使用命令编译这个简单的链表测试程序时

g++ -o SLLtest SLLtester.cpp intSLList.o

我收到错误:

SLLtester.cpp: In function ‘int main()’:
SLLtester.cpp:4:27: error: expected type-specifier
SLLtester.cpp:4:27: error: cannot convert ‘int*’ to ‘intSLList*’ in initialization
SLLtester.cpp:4:27: error: expected ‘,’ or ‘;’

我错过了一些简单的东西，但我不确定是什么。链表的标题和定义编译没有问题。包含三个文件。

//intSLList.hh
#ifndef INT_LINKED_LIST
#define INT_LINKED_LIST

class intSLList {
public:
 intSLList(){head=tail=0;}
 void Print();
 void AddToHead(int);
 void AddToTail(int);
 int RemoveFromHead();
 int RemoveFromTail();

protected:
 struct Node {
  int info;
  Node *next;
  Node(int e1, Node *ptr = 0) {info = e1; next = ptr;}
 } *head, *tail, *tmp;
 int e1;
};

#endif

和定义:

//intSLList.cpp
#include "intSLList.hh"
#include <iostream>

void intSLList::AddToHead(int e1){
 head = new Node(e1,head);
 if (!tail)
  tail = head;
}

void intSLList::AddToTail(int e1){
 if (tail) {
  tail->next = new Node(e1);
  tail = tail->next;
 }
 else
  head = tail = new Node(e1);
}

int intSLList::RemoveFromHead(){
 if (head){
  e1 = head->info;
  tmp = head;
  if (head == tail)
   head = tail = 0;
  else
   head = head->next;
 delete tmp;
 return e1;
 }
 else
  return 0;
}

int intSLList::RemoveFromTail(){
 if (tail){
  e1 = tail->info;
  if (head == tail){
   delete head;
   head = tail = 0;
  }
  else {
   for ( tmp = head; tmp->next != tail; tmp = tmp->next);
   delete tail;
   tail = tmp;
   tail->next = 0;
  }
  return e1;
 }
 else return 0;
}

void intSLList::Print(){
 tmp = head;
 while( tmp != tail ){
  std::cout << tmp->info << std::endl;
  tmp = tmp->next;
 }
}

最后是主要功能:

#include "intSLList.hh"

int main(){
 intSLList* mylist = new intSLList::intSLList();
 for ( int i = 0; i < 10; i++ ){
  mylist->AddToTail(i);
 }
 mylist->Print();
}

感谢您的帮助。

最佳答案

intSLList* mylist = new intSLList::intSLList();

这是错误的。当我们编写 new intSLList() 时，我们不是在“调用构造函数”——只是命名类型——因此完全命名构造函数(如 intSLList::intSLList )是完全错误的。

所以:

intSLList* mylist = new intSLList();

无论如何，您在这里不需要动态分配:

#include "intSLList.hh"

int main()
{
   intSLList mylist;

   for (int i = 0; i < 10; i++) {
      mylist.AddToTail(i);
   }

   mylist.Print();
}

关于时间:2019-05-06 标签:c++error: expected type-specifier，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/27735032/