Question

嘿，我正在制作“制作自己的冒险游戏！”现在，除了必须通过洞游戏，我想制作一个作弊码系统，现在我试图宣布一个字符串女巫等于超过6个字我不明白是什么问题我只做了两个单词并没有错误，当我用2个单词做同样的事情时，我得到了错误。

Main.cpp | 27 | error：没有匹配函数来调用'std :: basic_string :: basic_string（const char [4]，const char [6]，const char [5]，const char [5]， const char [6]，const char [6]）'|

这是我的代码：

#include <iostream>
//LVL1
#include "C:\Users\QuestionMark\Desktop\Make Your Own Adventure\LVL1\Dog.h"
#include "C:\Users\QuestionMark\Desktop\Make Your Own Adventure\LVL1\Dream.h"
#include "C:\Users\QuestionMark\Desktop\Make Your Own Adventure\LVL1\GTFO.h"

using namespace std;

int Return();
int Continue();

int main(){

    cout << "Welcome to my 'MAKE YOUR OWN ADVENTURE GAME!!!'\n";
    cout << "Have Fun and enjoy the ride!\n";
    cout << "Would you like to put in a cheat code??\n";
    cout << "Yes or No, Cap Sensitive!\n";
    Return();
    return 0;
}

int Return(){
        std::string y("Yes","No");
        cin >> y;
if(y.compare("Yes")){
        cout << "Please Enter Cheat Code now\n";
        std::string z("Dog","Dream","GTFO","Path","Sword","Weird");
        cin >> z;
    if(z.compare("Dog")){
        Dog();
    }else if(z.compare("Dream")){
        Dream();
    }else if(z.compare("GTFO")){
        GTFO();
    }else if(z.compare("Path")){
        Path();
    }else if(z.compare("Sword")){
        Sword();
    }else if(z.compare("Weird")){
        Weird();
    }else{
    cout << "Invalid Cheat Code\n";
   }

}else if(y.compare("No")){
return Continue();
}else{
    cout << "Invalid Answer!\n";
    Continue();
}
}

int Continue(){

    cout << endl;
    cout << "You wake up and your house is on fire what do you do ??\n";
    cout << "Quick Grab The Dog = 0, GTFO = 1, Go back to sleep = any other number\n";
    int x;
    cin >> x;
    if(x == 0){
         Dog();
    }else if(x == 1){
         GTFO();
    }else{
         Dream();
    }
}

Answer 1

你的问题在于你的字符串z的声明，你的代码是：

std::string z("Dog","Dream","GTFO","Path","Sword","Weird");

你的编译器找不到std :: string的构造函数，它接受6个参数而不是

std::string z("any string");

或因为您即将初始化z只是

std::string z;

没有用于调用'std :: basic_string <char> :: basic_string c ++ </char>的匹配函数

1 个答案: