c - 对不同的数组使用指针？

Question

我无法弄清楚如何让此 Hangman 程序根据用户输入选择要使用的正确数组。虽然我意识到我可以轻松设置包含相同代码的不同 if 语句，但这似乎是不必要的内存使用量。所以我的问题是:我可以使用指针(或者最好是更简单的东西)来表示数组名称，这样我只需要一个代码序列吗？我尝试通过指针来完成此操作，但我相信我弄错了，所以如果有人有任何提示，我将不胜感激。

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


int main(){

char easy_animals[2][3] = {
        { 'c', 'a', 't' }, //0
        { 'd', 'o', 'g' }, //1
      };

char easy_names[2][3] = {
        { 'p', 'a', 't' }, //0
        { 'b', 'o', 'b' }, //1
      };

char u,
    newline,
    dis[16] = { '_', '_', '_', '_' };
    input[10];

int random,
    guesses = 3,
    finish = 0;

_Bool successfulGuess = false;

srand(time(NULL));
random = rand() % 13;

printf("Animals or names?\n");
gets(input);

if (input[0] == 'a'){ // Or any other letter to signify the correct subject

    printf("Animal %d\n", random); // Check random number
    printf("---------\n\n");

    while (guesses > 0){
        finish = 0;
        successfulGuess = false;
        printf("Enter a letter: ");
        u = getchar();
        newline = getchar();

        for (int i = 0; i < 3; i++){

            if (u == dis[i]){
                successfulGuess = true;
                printf("\nYou already guessed this letter.\n");
                printf("\ninput = dis[i]\nGuesses left: %d\n\n", guesses);
                break;
            }
            else if (easy_animals[random][i] == u){
                successfulGuess = true;
                dis[i] = u;
                printf("\ninput = array char\nGuesses left: %d\n\n", guesses);
            }
        }
        printf("\n");

        for (int i = 0; i < 3; i++){
            printf("%c", dis[i]);
        }

        if (successfulGuess == false){
            guesses--;
            printf("\n\nbool statement\nGuesses left: %d\n\n", guesses);
        }

        if (guesses == 0){
            printf("Sorry, you've run out of guesses.");
        }

        for (int i = 0; i < 3; i++) {
            if (dis[i] != '_') {
                finish++;
            }
            if (finish == 3){
                printf("\n\nYou guessed the word!");
                guesses = 0;
            }
            else{
                continue;
            }
        }
        printf("\n\n");
    }

}
system("pause");
}

Answer 1

easy_animals 和 easy_names 都是指向 char 的指针数组。选择用户必须猜测的单词只意味着选择这些指针之一，因此请声明自己一个可用于表示该单词的 char*:

char* wordToGuess;

然后将其分配给从相关数组中随机选择的指针，如下所示:

if (/*user selects "animals"*/)
{
    wordToGuess = easy_animals[random];
}
else if (/*user selects "names"*/)
{
    wordToGuess = easy_names[random];
}
else (/*input is invalid*/)
{
    // Prompt the user to re-enter his/her choice.
}

最后，在确定用户是否成功猜测时使用新指针:

else if (wordToGuess[i] == u){
    successfulGuess = 1;
    dis[i] = u;
    printf("\ninput = array char\nGuesses left: %d\n\n", guesses);
}