Question

我写了这段代码（我刚刚开始编码btw），并且在13:17发生错误：

错误：将'const char [2]'赋值给'char [1]'的不兼容类型

这个错误也可以在27:25找到。

这不是我如何修复它的问题，而是解释为什么使用if(opinion="y")不起作用，因为它是一个字符。我已经尝试使用cin.getline()而没有任何结果（我还没有学到（我确实使用#incude <string>，即使std库中包含字符串）

有什么想法吗？

#include <iostream>
#include <string>
using namespace std;
int main(){
    int mycarrots=10;        //Initiates number of carrots that I have (mycarrots)
    int yourcarrots=0;      //Inititate the numbe of carrots the user has
    int wantcarrots=0;      //initiates the number of carrots user wants
    int wantcarrots2=0;     //initiates how many more carrots user wants apart from the ones that I can give him
    char opinion1[1], opinion2[1];          //initiates opinion whether user want carrots. Opionion2 Initiates opinion whether user has enough money to buy more carrots
    int ymoney=0;           //initiates how much money the user has
    cout<<"I have some carrots I want to give away, would you like some? (y/n)"<<endl;          //initiates convo, ask user whether he wants carrts
    cin>>opinion1;                                                                              //input of opinion (y/n)
    if (opinion1="n"){                                                                       //if the opinion is no, execute "Have a good day"
        cout<<"Have a good day!"<<endl;
        }
    else {                                                                                      //otherwise, resume convo
        cout<<"How many do you want?"<<endl;
        cin>>wantcarrots;
        if (wantcarrots>mycarrots){
            cout<<"I don't have that many carrots, you'll have to get some from the store."<<endl;
            cout<<"They're $1.5 each, so you'll have to pay "<<(wantcarrots-mycarrots)*1.5<<" dollars";
            cout<<"do you have enough money for that? (y/n)"<<endl;
            cin.getline(opinion2,1);
            wantcarrots2=wantcarrots-mycarrots;
            if (opinion2="y"){
                      cout<<"I can give you "<<mycarrots<<" carrots, but you'll have to get the other"<<mycarrots-wantcarrots<<" from the store."<<endl;
                     cout<<"Now off you go to the store then.";
                     }
            else {
                     cout<<"how much money do you have?"<<endl;
                     cin>>ymoney;
                     if (ymoney>=(wantcarrots2*1.5)){
                        cout<<"Off you go to the store."<<endl;
                        }
                     else if (ymoney<(wantcarrots2*1.5)){
                        cout<<"You'll have to settle for "<<ymoney/1.5<<" carrots."<<endl;
                        }       
                    }}
            else{
                cout<<"fatal errors. i am not prgrammed to do this"<<endl;}
            }
       else {
            cout<<"Here are your "<<wantcarrots<<" carrots"<<endl;
            cout<<"now you have "<<yourcarrots<<" carrots"<<endl;}
return 0;
            }

Answer 1

if(opinion="y")无法正常工作，因为opinion未声明且if(opinion1="y")无法正常工作，因为您无法分配到数组。

要检查所阅读的内容是y，请尝试if(*opinion1 == 'y')。

积分

做比较，而不是分配。
在比较之前取消引用从数组转换的指针。您也可以使用opinion1[0]。
使用字符文字而不是字符串文字。

更新：正如@SamVarshavchik建议的那样，此代码将导致越界访问。

如果您不更改opinion1的类型，请使用cin.get(*option1);代替cin>>opinion1;

Answer 2

除了在其他答案中指出的此代码的问题之外，我还要指出，如果输入不是空字符串，则会导致内存损坏和未定义的行为。

char opinion1[1]

一个字符数组只能容纳一个空字符串......

cin>>opinion1;

...只要在这里输入一个字符，就会溢出数组缓冲区，破坏堆栈，导致未定义的行为。

'const char [2]赋值为char [1]的不兼容类型

2 个答案: