没有匹配函数调用 'std：：basic_string<char>：：basic_string C++

Question

嘿，我正在制作一个"制作你自己的冒险游戏！"现在，除了必须在洞里玩游戏之外，我想制作一个作弊代码系统，现在我正试图宣布一个字符串女巫等于6个以上的单词。我不明白问题是什么。我只用了两个单词就完成了，没有错误。然而，当我用超过2个单词做了同样的事情，却出现了错误。

Main.cpp|27|错误：调用"std:：basic_string:：basic_string（const char[4]，const char[6]，const char[5]，const char[5]，constr char[6]，const char[6]）"|时没有匹配的函数

这是我的代码：

#include <iostream>
//LVL1
#include "C:UsersQuestionMarkDesktopMake Your Own AdventureLVL1Dog.h"
#include "C:UsersQuestionMarkDesktopMake Your Own AdventureLVL1Dream.h"
#include "C:UsersQuestionMarkDesktopMake Your Own AdventureLVL1GTFO.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 nown";
        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 Coden";
   }
}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 numbern";
    int x;
    cin >> x;
    if(x == 0){
         Dog();
    }else if(x == 1){
         GTFO();
    }else{
         Dream();
    }
}

Answer 1

您的问题在于字符串z的声明，您的代码是：

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

编译器找不到需要6个参数的std:：string的构造函数，请尝试

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

或者因为你即将初始化z只是

std::string z;