擲骰子游戲未返回正確的值

Question

自上次以來大家大家都在這里找到了極大的幫助，我會再問一個問題

我的代碼沒有返回正確的值：play_game函數中出了點問題，我無法弄清楚它是什么。我相信所有情況都被涵蓋了，但最終還是搞砸了。 第二場比賽結束后，我每次玩游戲時代碼也不會循環播放。 這不是一項任務

有什么建議嗎？

#include <stdio.h>
#include <time.h>
#include <stdlib.h>

static int sum, point, win = 0, roll = 0;    
bool play_game(void);
int roll_dice(void);

int main(void){

srand(time(NULL)); 


play_game();

char input[10];

do{ point = 0;
    play_game();
    if(win == 1){ // I'm assuming that play returns whether you won or not
        printf("You won!\n");
    }else{
        printf("You lost!\n");
    }
    printf("Would you like to continue? y/n\n");
    gets(input);
}while(*input == 'y'); // gets() flushes the buffer for next time you need input
return 0;
}


bool play_game(void){

point=0;
roll_dice();
printf("Your point is %d\n", sum);

while(roll == 1) /* first round */
{
  if(sum == 7 || sum == 11)
     return win = 1;
  else if(sum == 2 || sum == 3 || sum == 12)
     return win = 0;
  else if(sum == 1 || sum == 4 || sum == 5 || sum == 6 || sum == 8 || sum == 9 || sum    == 10){
     point=sum;
     roll_dice();
     }

}

while(roll > 1) /* all others rounds*/
{  
      if(sum == 7)
        return win = 0;
      else if(sum == point)
        return win = 1;
      else if(sum != point || sum != 7)
      roll_dice();

} 

}

int roll_dice(void){

int a,b;

a=1+rand() % (6);
b=1+rand() % (6);
sum=a+b;
roll++;
printf("You rolled %d\n", sum);
return sum;

}

輸出值

Answer 1

要點：

• 您可能需要1 + rand() % 6
• printf()的返回值可能不是您要從roll_dice()返回的值

Answer 2

循環需要更像是：

main(){
    char input[10];

    do{
        score = 0; //Always initialize the score
        if(play_game()){ // I'm assuming that play returns whether you won or not
            printf("You won!\n");
        }else{
            printf("You lost!\n");
        }
        printf("Would you like to continue? y/n\n");
        gets_s(input, 9);
    }while(*input == 'y'); // gets() flushes the buffer for next time you need input
}

Answer 3

Kyle的回答很好（如我所見），但是我可以發現一些問題，希望它能在更多情況下為您提供幫助。

• 您總是贏，我知道這很好，但是我敢打賭這不是預期的行為：

---

while(true) // This will always happen, because true is always evaluated as true
 {
  printf("Won\n\n");
  printf("Play again? y/n: ");
  break;
  }  

while(false) //This will never happen, since false is always evaluated as false
 {
  printf("Lost\n\n");
  printf("Play again? y/n: ");
  break;
  }

我認為您打算檢查play_game()的結果。 因此，添加另一個變量並對其進行檢查：

bool win;
win = play_game();
while (win == true)
...
while (win == false)
...

---

• 為什么在那里使用while循環？ 您還是會在第一次迭代中將其破壞

---

if(win == true)
{
  printf("Won\n\n");
}  
else
{
  printf("Lost\n\n");
}
printf("Play again? y/n: ");

---

• 該游戲將運行不超過兩次，因為您沒有一個循環取決於答案，而只有一個if語句僅被評估一次：

---

if(v=getchar() == 'y') //This is the second time the code runs, after that? nada.
 {
  point =0; /* reset point var */
  play_game();
  }
 else if(v=getchar() == 'n') // Why adding this check? you're going out anyway after the if-else
  exit(1);

---

編輯

當您使用while循環時，您的意思是：
當（括號中的某些表達式）為true時，執行塊{..}中的代碼，然后再次檢查括號中的表達式。

如果編寫while(true) ， while true is true, execute the code in the block實際上編寫while true is true, execute the code in the block 。 這將永遠發生。
如果您編寫while(false)時實際編寫了while false is true, execute the code in the block 。 這個false永遠不會是真的，也永遠不會執行代碼塊中的代碼。
如果您想在此處獲得實際條件，可以使用while(play_game()) 。 這就像寫while the returned value from the function play_game is true, execute the code in the block一樣， while the returned value from the function play_game is true, execute the code in the block的代碼，然后僅當play_game函數返回true（表示游戲中獲勝）時才while the returned value from the function play_game is true, execute the code in the block 。

有很多不錯的C教程，從此處或此處開始

Answer 4

在您的游戲序列中，擲骰子的時間不正確。

#include <stdio.h>
#include <time.h>
#include <stdlib.h>

// add defines to make states easier to read
#define WIN 1
#define LOSE 0

static int sum, point, win = 0, roll = 0;    
//bool play_game(void); 
int play_game(void); // changed return type to be int
int roll_dice(void);

int main(void){

    srand(time(NULL)); 


    // play_game(); // unncessary

    char input[10];

    do
    {
        point = 0;
        //play_game();
        // if(win == 1){
        if(play_game()){ // use return value from play_game()
            printf("You won!\n");
        }else{
            printf("You lost!\n");
        }
        printf("Would you like to continue? y/n\n");
        // gets(input);
        fgets(input, sizeof(input), stdin); // a safer input read
    } while(*input == 'y'); // gets() flushes the buffer for next time you need input
    return 0;
}


// bool play_game(void)
int play_game(void) // changed return type to be int
{

    point=0;
    // remove as this messes up the roll sequence.
    // roll_dice();
    // incorrect place to display this message
    //printf("Your point is %d\n", sum);

    // the while loop here is unnecessary
    //while(roll == 1) /* first round */
    //{
        roll_dice(); // add for initial come out roll.
        if(sum == 7 || sum == 11) { // I use braces to remove ambiguity
            // return win = 1;
            return WIN;
        } else if(sum == 2 || sum == 3 || sum == 12) {
            //return win = 0;
            return LOSE;
        }
        // sum will never be 1
        // on that note if it control reaches here it will be one of the other numbers.
        //} else if(sum == 1 || sum == 4 || sum == 5 || sum == 6 || sum == 8 || sum == 9 || sum == 10){
        // point=sum;
        // roll_dice(); // remove as this messes up the roll sequence.
        // }
        point=sum;
        printf("Your point is %d\n", sum);

    //}

    // while(roll > 1) /* all others rounds*/
    while (1) // might as well loop forever 
    {  
        roll_dice(); // add for subsequent dice rolls
        if(sum == 7) {
            //return win = 0;
            return LOSE;
        } else if(sum == point) {
            // return win = 1;
            return WIN;
        }
        // remove as this is unnecessary
        //  else if(sum != point || sum != 7)
        // remove as this messes up the roll sequence.
          //roll_dice();

    } 

}

int roll_dice(void){

    int a,b;

    a=1+rand() % (6);
    b=1+rand() % (6);
    sum=a+b;
    // roll++; // unncessary
    printf("You rolled %d\n", sum);
    return sum;

}

Answer 5

從您的描述中很難分辨（請說出您期望發生什么，以及發生了什么），但是我注意到的第一件事是您正在為a和b滾動5面骰子。