分割故障(核心转储)C

segmentation fault (core dump) c++

本文关键字:转储 核心 故障 分割      更新时间:2023-10-16

我一直在为我的课堂分配编写代码,当我用g 命令将其刻录时,它给了我细分故障。该代码很大,但第一部分,当我插入MAP的元素时,它可以起作用(因为它写下了命令,然后压碎了),然后当它必须施放(这是带有重新定义操作员的分配)时,它就会崩溃。所以有人可以帮我。

#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;
}

我的接口:

#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);


};

一个问题立即是您没有从声明的函数返回值返回值的值:

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????
}

不从声明返回值的函数返回值是不确定的行为。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)您的输出流是f,不应是cout

2)您应该通过const reference

通过Posiljka

3)注意现在我们返回f

此功能可能遇到相同的问题:

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

在该功能中,您确实具有return语句,但并非所有路径都返回一个值。