[英]Does not name a type error
这是我遇到的最愚蠢的小问题,我找不到原因。
我知道已经问过很多错误,但是我读到的每个错误都是关于声明的顺序的,但是,我在函数中使用struct之前先声明了我的struct,但仍然得到该错误。
这是头文件:
#ifndef GRAPH_H
#define GRAPH_H
#include <iostream>
#include <string>
using namespace std;
class Graph {
public:
struct Room;
// destructor
~Graph();
// copy constructor
Graph(const Graph &v);
// assignment operator
Graph & operator = (const Graph &v);
//Create an empty graph with a potential
//size of num rooms.
Graph( int num );
//Input the form:
//int -- numRooms times
//(myNumber north east south west) -- numRooms times.
void input(istream & s);
//outputs the graph as a visual layout
void output(ostream & s , string str );
//Recursively searches for an exit path.
void findPath( Room * start );
//Moves room N E S or W
void move( Room * room , string direction );
//inputs the starting location.
void inputStart( int start );
//Searches the easyDelete array for the room with the
//number "roomNumber" and returns a pointer to it.
Room * findRoom( int roomNumber );
struct Room
{
bool visited;
int myNumber;
Room *North;
Room *East;
Room *South;
Room *West;
};
private:
int numRooms;
int _index;
int _start;
Room ** easyDelete;
string * escapePath;
Room * theWALL;
Room * safety;
};
#endif
具体错误是:错误:“房间”没有命名类型,它在谈论我的
Room * findRoom( int roomNumber );
函数,应该返回指向“房间”的指针。 我尝试将结构的实际定义放在“结构室”中。 是,无济于事。
#include <iostream>
#include <string>
#include "Graph.h"
using namespace std;
...
Room * Graph::findRoom( int roomNumber )
{
...
}
您确定错误指向此.h文件吗? 不应有此错误...
如果您这样编写实现(在.cpp中)
Room * Graph::findRoom( int roomNumber );
它应该抱怨,因为Room是Graph的一部分:
Graph::Room * Graph::findRoom( int roomNumber );
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