在 C 中计算破折号以退出 While 循环

Question

我使用 while 循环制作了一个刽子手游戏。我想使用破折号的数量 == 0 来打破循环，这样玩家就知道他们赢了。这个词隐藏在“-”后面。我的词是编码，所以它有 6 个破折号。

当我运行程序时，对这个词的第一个猜测是准确的破折号数。但是在第二次猜测之后，以此类推，它将所有猜测中的所有破折号加在一起，而不仅仅是那一回合。到第三回合，它说我有 9 个破折号，因为这就是所有回合中剩下的数量。关于如何解决的任何想法？

char word[] = { "Coding" };
char dash[] = {"------"};
char guess;

int numguesses = 10; 
int count;
int loopcount;
int i;
int dashcount;

int main()
{
printf("Your Word Is: %s", &dash);
//printf("\nWord: %c", &dash);

loopcount = 10;
count = 1;
while (count <= loopcount)
{
    printf("\n\nPick a letter to guess: ");
    scanf(" %c", &guess);

    printf("\n\nYour Guess Was %c. ", guess);
    count = count + 1;

    if (guess == 'c')
    {
        dash[0] = 'c';
        printf("\n\n %s", dash);
    }
    else if (guess == 'o')
    {
        dash[1] = 'o';
        printf("\n\n %s", dash);
    }
    else if (guess == 'd')
    {
        dash[2] = 'd';
        printf("\n\n %s", dash);
    }
    else if (guess == 'i')
    {
        dash[3] = 'i';
        printf("\n\n %s", dash);
    }
    else if (guess == 'n')
    {
        dash[4] = 'n';
        printf("\n\n %s", dash);
    }
    else if (guess == 'g')
    {
        dash[5] = 'g';
        printf("\n\n %s", dash);
    }
    else
        printf("That Letter Is Not Apart of this Word");
    
    printf("\n\nYou are at Guess Number %d, Your word is at %s", count, dash);
    
    for (i = 0; dash[i]; i++)
    {
        if (dash[i] == '-')
        {
            dashcount++;
        }
    }
    printf("\n\nDoes it work ?: %d", dashcount);
}

if (count < numguesses)
{
    printf("\n\n\nYou Won the Game, Your Awesome!!");
}
else
{
    printf("\n\n\nI'm Sorry, the word was %s.", word);
}

printf("\n\n");
return 0;
}

Answer 1

问题在于您没有重置 dashcount 变量，因此每次它都从以前的值开始并在左侧添加破折号

for (i = 0; dash[i]; i++)
{
    if (dash[i] == '-')
    {
        dashcount++;
    }
}

您可以通过这种方式轻松解决此问题

dashcount=0;
for (i = 0; dash[i]; i++)
{
    if (dash[i] == '-')
    {
        dashcount++;
    }
}