Question

7   IntelliSense: no suitable user-defined conversion from "Difficulty" to "Difficulty" exists  e:\ICS3U CT\CT\game.h   29  10  CT

这是错误。下面列出了该文件的受影响部分。

 typedef struct Difficulty{
char* name;
char* description;
bool cumulatesHungerLoss;
int rngMod;
Difficulty(char* n, char* desc, bool cumulate, int rng){
name = n;
description = desc;
cumulatesHungerLoss = cumulate;
rngMod = rng;
}

};
Difficulty& easy = Difficulty("EASY","Normal RNG and Hunger Loss",false,0);
Difficulty& hard = Difficulty("HARD","Harder RNG and cumulative Hunger Loss",true,5);
Difficulty& insane = Difficulty("INSANE","Very Hard RNG and Cumulative Hunger Loss",true,15);
Difficulty getById(int id){
    switch(id){
    case 0:
        return easy;
    break;
    case 1:
        return hard;
    break;
    case 2:
        return insane;
    break;
    }
    return *NULL;
}

代码在C中，我只定义了一个名为Difficulty的结构类型。错误是返回容易;回归艰难;并且返回疯狂;。

每当我尝试将用户定义类型的值分配给该类型的另一个值时，就会发生这种情况。它搞乱了我的游戏，因为我需要将我的用户类型的值分配给数组并从函数返回值。这只影响我的一些类型。我可以保证我定义的每一种类型的唯一性。

[编辑]我打算用C编写代码。我正在编写的程序是一个C程序，如果它似乎是在C ++中，我很抱歉，它可能在C ++中解释，尽管意图是用C写的。

Answer 1

它是C ++，而不是C.我编辑了你的问题以解决这个问题。

Difficulty& easy = Difficulty("EASY","Normal RNG and Hunger Loss",false,0);
Difficulty& hard = Difficulty("HARD","Harder RNG and cumulative Hunger Loss",true,5);
Difficulty& insane = Difficulty("INSANE","Very Hard RNG and Cumulative Hunger Loss",true,15);

问题在于您是否尝试将临时对象绑定到非const引用。您不需要引用，只需直接定义一些命名对象：

const Difficulty easy("EASY","Normal RNG and Hunger Loss",false,0);
const Difficulty hard("HARD","Harder RNG and cumulative Hunger Loss",true,5);
const Difficulty insane("INSANE","Very Hard RNG and Cumulative Hunger Loss",true,15);

这不是唯一的问题，但它是你提出的问题。

这个错误是什么？它似乎没有理由出现

1 个答案: