Hangman游戏-C

Question

我正在做一个学校项目的绞刑者游戏，但我面临着一个问题：

do{
    system("cls");
    // Header of the game
    printf("\n HANGMAN GAME\n\n\n");

    // Present letters found
    for (i=0; word[i]!='\0'; i++)
        printf (" %c  ", word_2[i]);
        printf("\n");

    // Present positions to the letters
    for (i=0; word[i]!='\0'; i++)
        printf("___ ");
        printf("\n");

    // ****PLAYER'S ANSWERS*****

    // Read player's answers
    printf("\n\n Whats your guess + <enter>: ");
    scanf("%c", &letter);
    scanf("%c", &c);

    // Verify if the letter is in the word
    found=0;
    for(i=0; word[i]!='\0'; i++)
    if (word[i] == letter){
        word_2[i] = letter;
        corrects++;
        max_attemps--;
        found = 1;
        printf("\nWell done, %s. You have now %d attempts\n\n", name, max_attemps);
        system("pause");

    }
    if(found == 0){
        max_attemps--;
        printf("\nOh no, %s. You have now %d attempts\n\n", name, max_attemps);
        system("pause");
    }

    if (max_attemps <= 0 || corrects == lenght) {
        end = 1;
    }
} while (end == 0);

当我得到一个在单词中有两个或更多位置的正确字母时，它需要我进行两次或更多次尝试，因为system("pause")，而实际上我只能尝试一次。但是如果我不把system("pause")放进去，在我看到消息之前，板子就会被清除。有人知道我能做些什么来解决这个问题吗？我会非常感激的。

Answer 1

只有在扫描完整个单词之后，才能检查found标志：

    // Verify if the letter is in the word
found=0;
for(i=0; word[i]!='\0'; i++)
if (word[i] == letter){
    word_2[i] = letter;
    corrects++;
    max_attemps--;
    found = 1;

// comment out
//        printf("\nWell done, %s. You have now %d attempts\n\n", name, max_attemps);
//        system("pause");

}
    if(found == 1)
{
            printf("\nWell done, %s. You have now %d attempts\n\n", name, max_attemps);
            system("pause");
}



else{
    max_attemps--;
    printf("\nOh no, %s. You have now %d attempts\n\n", name, max_attemps);
    system("pause");
}

if (max_attemps <= 0 || corrects == lenght) {
    end = 1;
}