C++ 错误：预期的类型说明符
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/27735032/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
c++ error: expected type-specifier
提问by Michael LeVan
When I attempt to compile this simple linked list testing program using the command
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
I receive the error:
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 ‘;'
I am missing something simple but I am not sure what. The header and the definitions of the linked list compile without a problem. The three files are included.
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
And the definitions:
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;
}
}
And finally the main function:
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();
}
Thank you for the help.
感谢您的帮助。
回答by Lightness Races in Orbit
intSLList* mylist = new intSLList::intSLList();
This is wrong. When we write new intSLList(), we're not "calling the constructor" — merely naming the type — and therefore naming the constructor in full (as intSLList::intSLList) is flat-out wrong.
这是错误的。当我们编写 时new intSLList()，我们不是“调用构造函数”——只是命名类型——因此完全命名构造函数（as intSLList::intSLList）是完全错误的。
So:
所以：
intSLList* mylist = new intSLList();
You don't need dynamic allocation here anyway:
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();
}
回答by Some programmer dude
You're trying to call the constructor as a function. The constructor will be called automatically when you allocate the object, so change to
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，因此更改为
intSLList mylist;
That's it, no need for pointers or dynamic allocation at all.
就是这样，根本不需要指针或动态分配。
