C++开发：调用指定字符串数组元素时如何递增int变量？

当然可以实现！你当前的代码问题在于没有把DJ和他们对应的流派建立关联，而且注释里的尝试写法不符合C++语法（比如直接把字符串作为if条件、for循环格式错误）。下面我会给你一个清晰、可扩展的解决方案：

核心思路 #

1. 把DJ的名字和他们的主打流派绑定在一起（用结构体实现，方便后续扩展）
2. 建立流派名称到分数变量的映射，这样可以通过流派名直接找到对应的分数并递增
3. 根据用户的选择，遍历选中DJ的流派列表，给对应的分数加1

修改后的完整代码 #

#include "stdafx.h"
#include <iostream>
#include <string>
#include <vector>
#include <map>
using namespace std;

// 定义DJ结构体，绑定名字和对应的流派
struct DJ {
    string name;
    vector<string> genres;
};

int main() {
    cout << "Hello! This program is designed to figure out what Electronic Music you like based on artists presented and the answers you choose...\n" << endl;
    cout << "When you are ready to begin press \"Enter\"..." << endl;
    getchar();

    // 初始化所有流派的分数，初始值为0
    int bigRoom = 0;
    int deepHouse = 0;
    int drumBass = 0;
    int dubstep = 0;
    int electroHouse = 0;
    int futureHouse = 0;
    int hardDance = 0;
    int house = 0;
    int progressiveHouse = 0;
    int techno = 0;
    int trance = 0;
    int trap = 0;

    // 建立流派名称到分数变量的映射，方便通过字符串修改分数
    map<string, int&> genreScoreMap = {
        {"Big Room", bigRoom},
        {"Deep House", deepHouse},
        {"Drum & Bass", drumBass},
        {"Dubstep", dubstep},
        {"Electro House", electroHouse},
        {"Future House", futureHouse},
        {"Hard Dance", hardDance},
        {"House", house},
        {"Progressive House", progressiveHouse},
        {"Techno", techno},
        {"Trance", trance},
        {"Trap", trap}
    };

    // 初始化DJ列表，每个DJ对应他们的主打流派
    vector<DJ> djList = {
        {"DeadMau5", {"Progressive House", "Electro House", "Techno"}},
        {"Armin Van Buuren", {"Trance", "Progressive House", "Uplifting Trance"}},
        {"Avicii", {"Progressive House", "House", "Folk House"}},
        {"Ferry Corsten", {"Trance", "Progressive House", "Electro House"}},
        {"Kaskade", {"House", "Deep House", "Progressive House"}}
    };

    string userAnswer;
    cout << "Select the DJ you prefer by number. Otherwise select 3 if you don't know them. " << endl;
    // 暂时展示第2和第3位DJ（索引1和2），后续可以改成随机选择
    cout << "1 - " << djList[1].name << endl;
    cout << "2 - " << djList[2].name << endl;
    cin >> userAnswer;

    // 根据用户选择处理分数递增
    if (userAnswer == "1") {
        cout << "You have selected: " << djList[1].name << endl;
        // 遍历该DJ的所有流派，给对应分数加1
        for (const string& genre : djList[1].genres) {
            // 检查流派是否在映射中，避免无效流派导致错误
            if (genreScoreMap.find(genre) != genreScoreMap.end()) {
                genreScoreMap[genre]++;
            }
        }
    } else if (userAnswer == "2") {
        cout << "You have selected: " << djList[2].name << endl;
        for (const string& genre : djList[2].genres) {
            if (genreScoreMap.find(genre) != genreScoreMap.end()) {
                genreScoreMap[genre]++;
            }
        }
    } else if (userAnswer == "3") {
        cout << "You chose not to select either DJ." << endl;
    } else {
        cout << "Invalid selection! Please try again." << endl;
    }

    // 测试输出分数，验证功能是否正常
    cout << "\nCurrent Genre Scores:" << endl;
    cout << "Trance: " << trance << endl;
    cout << "Progressive House: " << progressiveHouse << endl;
    cout << "House: " << house << endl;

    return 0;
}

关键改进点说明 #

• 结构体绑定DJ与流派：用DJ结构体把每个DJ的名字和流派列表关联，后续添加新DJ只需在djList里新增元素，不用修改核心逻辑。
• 流派映射表：genreScoreMap让我们可以通过流派字符串直接找到对应的分数变量，避免写大量重复的if-else判断。
• 健壮性处理：增加了对无效输入和"都不选"选项的处理，同时检查流派是否存在于映射表中，避免程序崩溃。

后续随机选择DJ的扩展 #

你提到的随机选择DJ功能很容易实现，只需要用C++的<random>库生成两个不同的随机索引，然后展示对应的DJ即可，比如：

#include <random>
// ...
random_device rd;
mt19937 gen(rd());
uniform_int_distribution<> dist(0, djList.size()-1);
int djIndex1 = dist(gen);
int djIndex2 = dist(gen);
// 确保两个DJ不重复
while (djIndex2 == djIndex1) {
    djIndex2 = dist(gen);
}
// 展示随机选中的DJ
cout << "1 - " << djList[djIndex1].name << endl;
cout << "2 - " << djList[djIndex2].name << endl;

内容的提问来源于stack exchange，提问作者Matt