Question

我编写了这个程序来检查列表中给出的单词是否是彼此的字谜，并且由于某种原因，我的程序似乎只能运行一半，直到我得到一个运行时错误，表示＆＃34; map / set迭代器不可解除错误＆＃34;。我一直盯着这一段时间并试图调试它而我无法弄清楚问题，所以我希望有人在这里？以下是代码：

// Anagrams.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <string>
#include <fstream>
#include <vector>
#include <map>
#include <algorithm>
#include <set>


using namespace std;

int main()
{
    ifstream ifs("words.txt");// opens file
    if (!ifs) {
        cout << "no file";
        exit(1);
    }

    set<string>anagrams;
    map<string, string>words;
    string word;
    while (ifs >> word) {
        string sortedWord = word;
        sort(sortedWord.begin(), sortedWord.end());
        words.insert(pair<string, string>(word, sortedWord));
    }

    for (map<string, string>::iterator i = words.begin(); i != words.end(); i++) {
        //cout << "test i for loop" << endl;
        for (map<string, string>::iterator j = words.begin(); j != words.end(); j++) {
            //cout << "test j for loop" << endl;
            if (i->first == j->first) {
                j++;
            }
            if (i->second == j->second) {
                anagrams.insert(i->first);
                anagrams.insert(j->first);
                for (set<string>::iterator i = anagrams.begin(); i != anagrams.end(); i++) {
                    cout << *i << " ";
                }
                cout << endl;
                anagrams.clear();
                //cout << i->first << " is an anagram of " << j->first << " and their sorted version is " << i->second << endl;

            }
        }
    }
    return 0;
}

以下是单词列表：

monday
dynamo
abbot
acne
alert
alter
baby
Baby
BABY
best
bets
cane
later
Unix
UNIX

，在得到运行时弹出错误之前输出是这个：

acne cane
alert alter
alert later
alert alter
alert later
best bets
best bets
acne cane
dynamo monday
alert later
alert later
dynamo monday

然后是错误。

（请原谅我糟糕的输出，我可以稍后解决。现在更关注错误）

由于

map / set iterator no dereferencable似乎没有理由

0 个答案: