的指針訪問指針工會

Question

所以我有一個包含一個聯合結構如下：

struct FILL{ 
    char *name; 
    int id; 
}; 

struct TEST{ 
    union{ 
    struct FILL *fill; 
    int type; 
    } *uni; 
}; 

我不明白如何在結構中訪問工會成員。我一直試圖做到這一點如下：

struct TEST *test_struct, *test_int; 

test_struct = malloc(sizeof(struct TEST)); 
test_struct->uni = malloc(sizeof(struct TEST)); 
test_struct->uni->fill->name = NULL; 
test->struct->uni->fill->id = 5; 

test_int = malloc(sizeof(int)); 
test_int->uni->type = 10; 

但我得到segfaults，當我嘗試這個。我訪問這些錯誤嗎？我應該怎麼做呢？

編輯：對不起，我是專注于格式化和我搞砸了測試聲明。它已被修復。

Answer 1

每個該結構的指針構件必須被初始化，或者通過由malloc分配動態存儲，或分配給其他變量。這裏是你的代碼的問題：

struct TEST *test_struct, *test_int; 

test_struct = malloc(sizeof(struct TEST)); 
test_struct->uni = malloc(sizeof(struct TEST)); // uni should be allocated with size of the union, not the struct 
test_struct->uni->fill->name = NULL; // uni->fill is a pointer to struct FILL, it should be allocated too before accessing its members 
test->struct->uni->fill->id = 5; 

test_int = malloc(sizeof(int)); // test_int is of type struct TEST, you are allocating a integer here 
test_int->uni->type = 10; // same, uni not allocated 

所以請嘗試以下修正：

struct TEST *test_struct, *test_int; 

test_struct = malloc(sizeof(struct TEST)); 
test_struct->uni = malloc(sizeof(*test_struct->uni));   
test_struct->uni->fill = malloc(sizeof(struct FILL)); 
test_struct->uni->fill->name = NULL; 
test_struct->uni->fill->id = 5; 

test_int = malloc(sizeof(struct TEST)); 
test_int->uni = malloc(sizeof(*test_struct->uni));