使用不同數組的指針？

Question

我很難搞清楚如何讓這個Hangman程序根據用戶輸入選擇正確的數組。雖然我意識到我可以很容易地設置一個包含相同代碼的不同if語句，但這似乎是不必要的內存使用量。所以我的問題是這樣的：我可以使用指針（或者，最好是簡單些）來表示數組名稱，這樣我只需要一個代碼序列？我試圖通過指針做到這一點，但我相信我錯了，所以如果有人有任何提示，我會非常感激。

#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是指向char的兩個陣列。選擇一個單詞，用戶將不得不猜測僅僅意味着選擇這些指針之一，因此宣佈自己char*，你可以用它來表示字：

char* wordToGuess; 

然後從相關分配給隨機選擇的指針數組像這樣：

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