沒有匹配函數調用'std :: basic_string <char> :: basic_string C++

嘿，我正在製作一個「製作你自己的冒險遊戲！」現在，其他人不得不通過洞穴遊戲來製作一個作弊代碼系統，現在我試圖宣佈一個字符串女巫等於超過6個字我不明白問題是什麼我只用兩個並沒有任何錯誤，當我用超過2個單詞做同樣的事情時，我得到了錯誤。調用'std :: basic_string :: basic_string（const char [4]，const char [6]，const char [5]，const char [5]）時沒有匹配函數。，const char [6]，const char [6]）'|沒有匹配函數調用'std :: basic_string <char> :: basic_string C++

這裏是我的代碼：

#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(); 
    } 
}

來源

2013-11-29 Jesse Laviolette

我不知道你想做什麼，但查找在'的std :: string'參考。請注意'=='。 – chris

沒有帶六個參數的std :: string的構造函數。 – 2013-11-29 03:48:23

也沒有構造函數需要兩個字符串。你偶然碰到的那個是迭代器對構造函數，它應該會在程序運行時崩潰。 – chris

你的問題就在這裏爲你的字符串z中的聲明，你的代碼是：

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

你的編譯器不能找到性病構造::需要6個參數的字符串嘗試代替

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

還是因爲你將要初始化ž只是

std::string z;

來源

2013-11-29 04:23:07 Gibby

沒有匹配函數調用'std :: basic_string <char> :: basic_string C++

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