为什么我的超载不是<运算符不适用于STL排序 [英] Why isn't my overloading < operator not working for STL sort

问题描述

我有以下代码，我想根据字符串的最后一个字符对字符串向量进行排序.我已经完成了以下操作，但排序是默认规则完成的.

I have this following code where I want to sort vector of string according to the last character of the string. I've done the following but the sorting is done by default rules.

这是超载<部分:

Here's the overloading < part:

bool operator<(const string &s1, const string &s2){
    return s1.at(s1.size() - 1) < s2.at(s2.size() - 1);
}

这是主要内容:

vector <string> nameList;
int n;
cin>>n;
while(n--){
    string name;
    char str[100];
    cin>>str;
    name += str;
    nameList.push_back(name);
}

sort(nameList.begin(), nameList.end());
for(int i = 0; i < nameList.size(); i++)
    cout<<nameList.at(i)<<endl;

ideone中的代码: LINK

Code in ideone: LINK

推荐答案

如前所述，您的operator<没有被调用，std::string在std名称空间中已经有重载的运算符.

As noted, your operator< is not being called, std::string already has overloaded operators in the std namespace.

std::sort 有两个版本，一个将使用operator<和另一个使用自定义谓词对容器进行排序的对象. 添加自定义谓词，您仍然可以根据需要使用sort对vector进行排序；

There are two versions of std::sort, one that will use operator< and another that takes a custom predicate to sort the container. Add a custom predicate, you can still used sort to sort your vector as required;

#include <algorithm>
#include <string>
#include <iostream>
#include <vector>
using namespace std;

bool lastchar(const string &s1, const string &s2){
    return s1.at(s1.size() - 1) < s2.at(s2.size() - 1);
}

int main(){
        vector <string> nameList;
        int n;
        cin>>n;
        while(n--){
            string name;
            char str[100];
            cin>>str;
            name += str;
            nameList.push_back(name);
        }

        sort(nameList.begin(), nameList.end(), &lastchar);
        for(int i = 0; i < nameList.size(); i++)
            cout<<endl<<nameList.at(i);

        return 0;
}

我将其称为lastchar，但您可以将其命名为最好的.另外，如果您可以使用C ++ 11或更高版本，则可以使谓词为lambda.

I've called it lastchar, but you can name it what ever is best. As an added bonus, if you can used C++11 or above, you can make the predicate a lambda.

sort(nameList.begin(), nameList.end(), [] (const string &s1, const string &s2){
  return s1.at(s1.size() - 1) < s2.at(s2.size() - 1);
});

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