c++ error: expected type-specifier
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 ‘;’
我遗漏了一些简单的东西,但我不确定是什么。 header 和链表的定义编译没有问题。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
以及定义:
//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。
您正在尝试将构造函数作为函数调用。分配对象时会自动调用构造函数,所以改为
intSLList mylist;
就是这样,根本不需要指针或动态分配。
intSLList* mylist = new intSLList::intSLList();
这是错误的。当我们写 new intSLList()
时,我们不是 "calling the constructor"——只是命名类型——因此完整命名构造函数(如 intSLList::intSLList
)是完全错误的。
所以:
intSLList* mylist = new intSLList();
反正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();
}
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 ‘;’
我遗漏了一些简单的东西,但我不确定是什么。 header 和链表的定义编译没有问题。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
以及定义:
//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。
您正在尝试将构造函数作为函数调用。分配对象时会自动调用构造函数,所以改为
intSLList mylist;
就是这样,根本不需要指针或动态分配。
intSLList* mylist = new intSLList::intSLList();
这是错误的。当我们写 new intSLList()
时,我们不是 "calling the constructor"——只是命名类型——因此完整命名构造函数(如 intSLList::intSLList
)是完全错误的。
所以:
intSLList* mylist = new intSLList();
反正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();
}