stl map <>中的分段错误

Question

我不是很擅长c ++,但我必须测试一个基于声誉的系统.下面给出了一个代码片段,当我在ubuntu系统上运行它时,它给出了段错误.正如我在评论中写的那样,两个函数"tackleFirstHandInfo()"和"updateReputation()"分别正确运行但是当我从另一个函数调用一个函数时它会崩溃.任何帮助将不胜感激,提前感谢.代码如下:

"ex.h"

#ifndef _ex_h
#define _ex_h
#include "iostream"
#include <map>
#define FADING 0.9
enum Behaviour {FORWARDING, NOTFORWARDING};

class Rating
{

private:
    double reputation;

public:

    Rating() { reputation = 5.0; }
    Rating(double rep) {reputation = rep;}

    ~Rating() {}

    double getRep() { return reputation; }

    void updateRep(Behaviour behaviour) {
        if (behaviour == FORWARDING)
            reputation = reputation + 1;    
        else 
            reputation = reputation - 1;    
    } 

};

#endif

"ex.cc"

#include <map>                                  
#include <string>
#include <iostream>   
#include "ex.h"                           
using namespace std;
typedef map<int, Rating*> ratingTable;
class RepSys {
private:
    ratingTable repTable; 
    map<int, Rating*> fHandInfo;
    Rating* rating;

public: 
    RepSys(){}
    ~RepSys(){}

    void tackleFirstHandInfo(int address, Behaviour behaviour)
    /* This Function and the function below individually run correctly */
    {
        map<int, Rating*>::iterator it;
        it=fHandInfo.find(address);
        if (it == fHandInfo.end()) {
            cout << "Adding New Entry for (fHandInfo) "<< address <<endl;
            rating = new Rating();   
            fHandInfo[address] = rating;
        }
        (it->second)->updateRep(behaviour);
        cout<<"First Hand Reputation of "<<address<<"\t is ="<< (it->second)->getRep()<<endl;
        updateReputation(address, behaviour); // This causes SegFault !!!!
        return;
    }

    void updateReputation(int address, Behaviour behaviour)
    {
        map<int, Rating*>::iterator it;
        it = repTable.find(address);
        if (it == repTable.end()) {
            cout << "Adding New Entry for (repTable) "<< address <<endl;
            rating = new Rating(); 
            repTable[address] = rating;
        }
        (it->second)->updateRep(behaviour);
        cout<<"Reputation of "<<address<<"\t is ="<< (it->second)->getRep()<<endl;
    }   
};

int main() {
    int address;    
    RepSys repsys;
    while (address != 0)
    {                                
        cout << "Address\n";
        cin >> address;
        repsys.tackleFirstHandInfo(address, FORWARDING);
    }
    return 0;
}

Answer 1

您在两个函数中都遇到了一个主要问题,它是:

if (it == fHandInfo.end()){
    // Some code that doesn't alter 'it'
}
(it->second)->updateRep(behaviour);

如果it确实指向最后,则它不是可解除引用的,因此it->second具有未定义的行为.如果要插入某些内容并想要it指向它,则必须重做find或使用返回迭代器(或包含迭代器的对)的insert方法,并重新分配it给返回值的正确部分.

编辑

还有几点:

class RepSys {
private:
    ratingTable repTable; 
    map<int, Rating*> fHandInfo;
    Rating* rating;

你已经typedef编过ratingTable了map<int, Rating*>.将typedef用于一个类变量而不是另一个变量似乎有点不一致.

rating是一个类变量,但您似乎只将它用作两个函数中的临时持有者.如果这是您的预期用途,最好在两个函数中使用局部变量.

您永远不会放置地图中delete的Rating对象.如果映射应该拥有Rating对象,那么从对象生存期/内存管理的角度来看,更容易实现,std::map<int, Rating>这样您就不必进行任何手动删除.似乎没有Rating设计为基本调用,它是一个值类.