访问指向联合的指针中的指针

Question

所以我有一个包含联合的结构如下：

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

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

我不明白如何访问结构中的工会成员。我一直在尝试这样做：

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;

但是当我尝试这个时我得到了段错误。我访问这些错误吗？不然怎么办？

编辑：对不起，我专注于格式，我搞砸了 TEST 的声明。已经修好了。

Answer 1

结构的每个指针成员都必须初始化，要么通过malloc分配动态存储空间，要么分配给其他变量。以下是您的代码的问题：

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

所以尝试以下修复：

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));