结构定义中的预期 ' '

Question

我现在正在做一个学校项目，我需要定义两个结构作为地址，如下面的代码所示：

typedef struct board_t* board;
/**
 * @brief Pointer to the structure that holds the game.
 */

typedef struct piece_t* piece;
/**
 * @brief Pointer to the structure that holds a piece
 */

如果我让它喜欢它，它就会编译。但是，一旦我尝试用括号替换分号来定义结构，就会出现编译错误。这是代码和错误：

typedef struct piece_t* piece{
/**
 * @brief Pointer to the structure that holds a piece
 */
 enum shape p_shape;
 enum size p_size;
 enum color p_color;
 enum top p_top;
 enum players author;
};


typedef struct board_t* board{
/**
 * @brief Pointer to the structure that holds the game.
 */
 piece array[4][4];
}

错误：

error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘{’ token
   53 | typedef struct board_t* board{

我需要做的是创建一个板，其中包含我可以在内部函数中编辑的部分。谁能帮帮我？

Answer 1

我认为 typedef 名称需要放在最后

typedef struct piece_struct {
/**
 * @brief Pointer to the structure that holds a piece
 */
 enum shape p_shape;
 enum size p_size;
 enum color p_color;
 enum top p_top;
 enum players author;
}
piece;


typedef struct board_struct {
/**
 * @brief Pointer to the structure that holds the game.
 */
 piece array[4][4];
}
board;

如果您想要指针的 typedef 名称，则需要单独创建它们。

typedef piece* piece_ptr;
typedef board* board_ptr;

如果将结构定义与 typedef 分开，代码可能会更清晰：

struct piece_struct {
/**
 * @brief structure that holds a piece
 */
 enum shape p_shape;
 enum size p_size;
 enum color p_color;
 enum top p_top;
 enum players author;
};

typedef piece_str* piece;  // piece is a new name for a pointer
                           // to a piece_str

struct board_struct {
/**
 * @brief structure that holds the game.
 */
 piece array[4][4];
};

typedef struct board_struct* board;   // board is a new name for a pointer
                                      // to a board_str

我个人倾向于不为指针创建 typedef，因为我发现很难记住它是指针还是结构本身，所以我为结构创建了 typedef，并在声明指针时使用 *。

结构定义中的预期 ' '

Expected ' ' in definition of struct

c

struct

definition