C 返回结构数组时出现问题

我是个新手，我对类型结构的数组有问题。我真的不知道在哪里指定数组是指针，在哪里不指定它。所以，无论我在声明和调用涉及这个结构数组的函数时对指针做什么组合，它都不会编译。你知道为什么吗？ 以下是相关代码：

typedef struct{
  char letter[7];
} player;

player init_player(){
  player tab[2];

  char letter_player1[4]={' ','a','b','c'};
  char letter_player2[4]={' ','x','y','z'};
  strcpy(tab[1].letter,letter_player1);
  strcpy(tab[2].letter,letter_player2);

  return *tab;
}

void displaying(player tab){
  int position=1;
  printf("%c\n",*tab[1].letter[position]);
}

int main(){
  player tab=init_player(tab);
  displaying(tab);
}

---

提前谢谢你

您的代码如何出错：

/* required #include (or function prototypes) are not written */

typedef struct{
  char letter[7];
} player;

player init_player(){ /* the return type is not for returning arrays */
  player tab[2];

  /* no terminating null characters are placed, so cannot be used with strcpy() */
  char letter_player1[4]={' ','a','b','c'};
  char letter_player2[4]={' ','x','y','z'};
  strcpy(tab[1].letter,letter_player1);
  strcpy(tab[2].letter,letter_player2); /* only tab[0] and tab[1] are available, tab[2] is out-of-bounds */

  return *tab; /* not the array but only the first element of the array is returned */
}

void displaying(player tab){ /* the argument type is not for getting arrays */
  int position=1;
  /* even if tab were an array(pointer), dereferencing is done via [] operator, so you don't need * here */
  printf("%c\n",*tab[1].letter[position]);
}

int main(){
  player tab=init_player(tab); /* the variable type is not for dealing with arrays */
  displaying(tab);
}

如何修复：

/* add #include */
#include <stdio.h> /* for printf() */
#include <stdlib.h> /* for malloc(), free() and exit() */
#include <string.h> /* for strcpy() */

typedef struct{
  char letter[7];
} player;

/* change return type to return a pointer */
player* init_player(){
  /* you cannot use non-static local array, so instead dynamically allocate an array here */
  player* tab = malloc(sizeof(*tab) * 2);

  /* add terminating null characters (increase array size, and they will be automatically added) */
  char letter_player1[5]={' ','a','b','c'};
  char letter_player2[5]={' ','x','y','z'};

  if (tab == NULL) exit(1); /* for in case the allocation failed */
  /* use tab[0] and tab[1] instead of tab[1] and tab[2] */
  strcpy(tab[0].letter,letter_player1);
  strcpy(tab[1].letter,letter_player2);

  return tab; /* return the (pointer to the first element of) array */
}

/* change argument type to get a pointer */
void displaying(player* tab){
  int position=1;
  /* remove extra * (indirection operator) */
  printf("%c\n",tab[1].letter[position]);
}

int main(){
  /* change variable type to store a pointer */
  player* tab=init_player(tab);
  displaying(tab);
  free(tab); /* the array is dynamically allocated, so clean-up it */
}

您不需要将参数选项卡传递给init_player，

---

但是它是无害的，因为init_player可以接受任何参数，并且不使用这些参数。

strcpy需要一个以nul结尾的字符串：char letter_player1[]={'，'a'，'b'，'c'，'\0'}；选项卡[2]不存在，c中的数组基于0。最后，返回一个本地数组，其生存期以函数UB结束。@DavidRanieri注释的最后一部分是错误的。返回的不是本地数组，而是本地数组的第一个元素，因此没有UB用于返回将变为无效的指针。@MikeCAT在function:main中，语句：player*tab=init_playertab；这是不对的。因为函数：init_player没有参数。@MikeCAT哇谢谢你这么全面的回答，我没想到会有这么多错误！有办法绕过马洛克吗？如果不是，我想是时候学习了哈哈