Question

有人知道这段代码有什么问题。它会带来上面的错误。我正在尝试创建一个生成“随机”字母的程序，并为用户提供最多6次猜测该字母的机会。

if ( Play != "y" || Play != "Y" )

编辑：完整代码

// This function displays game instructions, and returns nothing.
void Instructions();

// This function plays one game, and returns "W" if the player wins
// or "L" if the player runs out of guesses. 
char Play();

//this function prompts the player to make a guess and returns that guess
char getLetter();

//The function returns 1 if the guess matches the solution and returns a 0 if they do not match
char guess();

// This function returns a random letter between "A" and "Z 
char getAnswer();

int CompareLetters(char guess, char answer);

int main()
{ 

    char answer;


    //1. Greet the user and ask if they would like to play a guessing game.
    printf("\nHello there!\nWould like to play a guessing game?Enter Y or N: \n");
    scanf(" %c", &answer);

    if(answer == 'y' || answer == 'Y')
        Instructions();
    {
        printf("\nYou entered Y.  Let's  play!\n");

        do{

        }while (answer == 'y' || answer == 'Y');
    }
    printf("\nMaybe next time.\n");
    printf("\nGoodBye for now.\n");
    return -1;
} 


void Instructions()
{
    printf("I have a capital letter in mind. You have 6 chances to guess which letter I am \nthinking. I will let you know if you are too high or too low.\n");
    printf("After each guess, you will be informed if your guess is too high or too low.\nGood luck!\n");

}

int PlayGuess(char answer)
{
    int NumGuesses=0; int WinOrLose=0;

    while (NumGuesses < MAX_GUESSES && WinOrLose==0);
    {

        //6. If the player guesses wrong for a 6th time, console them and let the program end with a return code of 1.  
        char guess;
        guess = getLetter();
        CompareLetters(guess,answer);
        if(CompareLetters(guess,answer)==1)
        {
            WinOrLose = 1;
        }
        else
        {
            NumGuesses++;
        }
    }
    if (WinOrLose==1)
    {
        return 1;
    }
    else
    {
        return 0;
    }
}

//3. Generate a "random" character between 'A' and 'Z'.  This will be the value the player will try to guess. 
char getLetter()
{
    char guess=0;
    printf("Please enter your letter guess:", guess);
    scanf(" %c",&guess);
    return guess;
}

int CompareLetters(char guess, char answer)
{
    if(guess == answer)
    {
        return 1;
    }
    else
    {
        ////5. If the player does not guess the right answer, display whether the guess is "too high" or "too low". 
        if (guess < answer)
        {
            printf("Your guess is too low.");
        }
        else
        {
            printf("Your guess is too high.");
        }

        {
            printf("\nDo you want to play again? Y or N: \n");

        }
        if ( Play != 'y' &&  Play != 'Y' )
            printf("Thanks for playing.Goodbye!/n");
}

Answer 1

Play似乎是（单个）char，而不是char数组。因此，您应该使用'y'代替"y"。

if ( Play != 'y' || Play != 'Y' )

这里似乎也存在逻辑错误 - 大概你打算使用&&而不是||：

if ( Play != 'y' && Play != 'Y' )

编辑：您现在已经添加了其余的源代码。这里有一个很多的问题，但我会列出一些让你入门。

您对Instructions()的调用不在大括号内，因此将有条件地调用Instructions()，其余代码将无条件执行。

if(answer == 'y' || answer == 'Y')
    Instructions();
{
    printf("\nYou entered Y.  Let's  play!\n");

这应该是：

if(answer == 'y' || answer == 'Y')
{
    printf("\nYou entered Y.  Let's  play!\n");
    Instructions();

您的do {} while声明为空。它也是一个无限循环，因为答案的scanf不在循环之内。
```
    do{

    }while (answer == 'y' || answer == 'Y');
```
您应该将scanf 移到循环中，然后添加对Play()的调用。
与Play中的CompareLetters进行比较的if语句是不正确的（没有名为Play的变量）并且位置错误（CompareLetters不应该对此负责。）

您应该将其移至main，更新并与我上面提到的answer变量进行比较。

Answer 2

在C中，双引号中的字符序列表示一个字符串，在内存中为const char *。您的变量Play包含一个char，而不是char*。告诉编译器使用char常量的方法是在字符周围使用单引号，如下所示：

Play != 'Y'

Answer 3

变量Play正在使用char，因此请尝试使用''。此外，""用于char * literal。您可以尝试这样：

if ( Play != 'y' || Play != 'Y' )

而不是

 if ( Play != "y" || Play != "Y" )

不兼容的操作数类型（'char'和'char *'）？

3 个答案: