Closed. This question is off-topic。它当前不接受答案。
                            
                        
                    
                
                            
                                
                
                        
                            
                        
                    
                        
                            想改善这个问题吗? Update the question,所以它是on-topic,用于堆栈溢出。
                        
                        3年前关闭。
                                                                                            
                
        
我一直在为类分配编写代码,当我使用g ++命令在终端中发掘代码时,它给我分段错误。代码很大,但是第一部分,当我插入map元素时,它可以工作(因为一旦编写了命令然后被压碎了),然后当它必须进行强制转换(这是Redefinition运算符的赋值)时,它便崩溃了。有人可以帮我吗

#include <iostream>
#include <map>
#include <algorithm>
#include <string>
#include <typeinfo>
#include "druga_verzija.h"

using namespace std;
//constructor
Posiljka::Posiljka(){
    m.clear();

}


//how to print my map
ostream&operator<<(ostream&f, Posiljka&p){
    map<string , int>::iterator it;
    it=p.m.begin();
    cout << it->second << "x" << it->first;
    for( it=p.m.begin() ; it != p.m.end() ; ++it){
        cout << " " << it->second << "x" << it->first;
    }
}

//how to insert elements
Posiljka&Posiljka::operator<<(string s){
    if(m.empty()){
        m.insert(pair<string,int>(s,1));
        return *this;
    }
    map<string ,int>::iterator it=m.begin();
    for(; it != m.end() ; ++it){
        if( it->first == s){
            it->second++;
            return *this;
        }
        if( it->first > s){
            m.insert( it, pair<string,int>(s , 1));
            return *this;
        }
    }
    if(it == m.end()){
        m.insert(pair<string,int>( s, 1));
        return *this;
    }
}

//how to delete them
Posiljka&Posiljka::operator>>(string s){
    if(m[s]){
        m[s]=m[s]-1;
        if(!m[s]){
            m.erase(s);
            return *this;
        }
        return *this;
    }
    return *this;
}
//how to make new map that contains two old maps
Posiljka Posiljka::operator|(Posiljka &p){
    Posiljka novi;
    map<string,int>::iterator it;
    for( it = p.m.begin() ; it!= p.m.end() ; ++it)
        novi.m[it->first]=it->second;
    for( it = m.begin(); it != m.end() ; ++it){
        if(novi.m[it->first])
            novi.m[it->first] = novi.m[it->first] + it->second;
        else
            novi.m[it->first] = it->second;
        }
    return novi;
}

//multiply map witn int and make new map
Posiljka Posiljka::operator*(int x){
    map<string , int>::iterator it;
    Posiljka novi;
    for( it=m.begin() ; it!=m.end() ; ++it)
        novi.m[it->first] = it->second * x;
    return novi;
}

//write how much objects map has
Posiljka::operator int(){
    int suma=0;
    map<string , int>::iterator it;
    for( it = m.begin() ; it != m.end() ; ++it)
        suma = suma + it->second;
    return (int)suma;
}
//write maks of one object
int Posiljka::operator+(){
    int maks=0;
    map<string, int>::iterator it;
    for( it = m.begin() ; it != m.end() ; ++it)
        if( it->second > maks)
            maks=it->second;
    return maks;
}

//write min of one object
int Posiljka::operator-(){
    if(m.begin()->second)
        return 0;
    int mini=m.begin()->second;
    map<string, int>::iterator it;
    for( it=m.begin() ; it!=m.end() ; ++it)
        if( it->second < mini)
            mini=it->second;
    return mini;
}
//if there is object s in thios map
bool Posiljka::operator()(string s){
    if(m[s])
        return true;
    else
        return false;
}


我的主要:

#include <iostream>
#include "druga_verzija.h"
using namespace std;
int main()
{
Posiljka P, Q, R;
P << "olovka" << "tipkovnica" << "olovka" << "olovka";
cout << P << endl;
P << "olovka" << "monitor" << "tipkovnica" << "gitara";
cout << P << endl;
//1xgitara 1xmonitor 4xolovka 2xtipkovnica
cout << (int)P << endl;
//8
Q = P*2;
cout << Q << endl;
//2xgitara 2xmonitor 8xolovka 4xtipkovnica
R = P | Q;
cout << R << endl;
//3xgitara 3xmonitor 12xolovka 6xtipkovnica
cout << -R << " " << +R << endl;
//3 12
while ( R("olovka") )
R >> "olovka";
cout << R << endl;
//3xgitara 3xmonitor 6xtipkovnica
R >> "gitara" >> "monitor" >> "tipkovnica" >> "tipkovnica";
cout << R << endl;
//2xgitara 2xmonitor 4xtipkovnica
return 0;
}


我的界面:

#include <iostream>
#include <map>
#include <string>
#include <typeinfo>


using namespace std;

class Posiljka
{
private:

map<string,int> m;

public:

Posiljka();
friend ostream&operator<<(ostream&, Posiljka&);
Posiljka&operator<<(string s);
Posiljka&operator>>(string s);
Posiljka operator|(Posiljka &p);
Posiljka operator*(int x);
operator int();
int operator+();
int operator-();
bool operator()(string s);





};

最佳答案

立即出现的一个问题是,您不是从声明为返回值的函数中返回值:

ostream&operator<<(ostream&f, Posiljka&p)
{
    map<string , int>::iterator it;
    it=p.m.begin();
    cout << it->second << "x" << it->first;
    for( it=p.m.begin() ; it != p.m.end() ; ++it)
    {
        cout << " " << it->second << "x" << it->first;
    }

   // Where is the return????
}


从声明为返回值的函数中不返回值是未定义的行为。该功能可能应该是:

ostream&operator<<(ostream &f, const Posiljka &p)
{
    map<string , int>::const_iterator it;
    it=p.m.begin();
    f << it->second << "x" << it->first;
    for( it=p.m.begin() ; it != p.m.end() ; ++it)
    {
        f << " " << it->second << "x" << it->first;
    }
    return f;
}


1)您的输出流为f,不应为cout

2)您应通过const reference通过Posiljka。

3)请注意,现在我们返回f

此功能可能会遇到相同的问题:

Posiljka&Posiljka::operator<<(string s)

在该函数中,您确实有return语句,但并非所有路径都返回值。

10-06 16:16