不兼容類型的分配 '爲const char [2]到char [1]'

Question

我寫了這個代碼（我剛開始編碼BTW），並有一個錯誤在13:17：

error: incompatible types in assignment of 'const char [2]' to 'char [1]'

此錯誤也可以在27:25找到。

這不是我如何修復它的問題，而是解釋爲什麼使用if(opinion="y")不起作用，因爲它是字符。我已經使用cin.getline()沒有任何結果（我還沒有得知，但（我沒有使用#incude <string>即使字符串包含在std庫）

任何想法試過嗎？

#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")不會工作，因爲opinion未聲明和if(opinion1="y")將無法​​正常工作，因爲你不能分配給數組。

要檢查什麼ISS讀爲y，嘗試if(*opinion1 == 'y')。

積分是

• 做比較，而不是分配。
• 取消引用在比較之前從數組轉換的指針。您也可以使用opinion1[0]。
• 使用字符而不是字符串文字。

UPDATE：作爲@SamVarshavchik表明，此代碼將導致出界外的訪問。

如果不改變的opinion1類型，使用的cin.get(*option1);代替cin>>opinion1;

Answer 2

除了與該代碼多數民衆贊成指出，在其他的答案的問題，我還將指出，如果輸入的是什麼，除了空字符串之外，這會導致內存損壞和未定義的行爲。

char opinion1[1] 

一個字符數組只可容納一個空字符串...

cin>>opinion1; 

...並儘快在此處輸入一個字符，這將溢出數組緩衝區，損壞堆棧，並導致未定義的行爲。