的指针访问指针工会

Question

所以我有一个包含一个联合结构如下：

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

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

我不明白如何在结构中访问工会成员。我一直试图做到这一点如下：

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，当我尝试这个。我访问这些错误吗？我应该怎么做呢？

编辑：对不起，我是专注于格式化和我搞砸了测试声明。它已被修复。

Answer 1

每个该结构的指针构件必须被初始化，或者通过由malloc分配动态存储，或分配给其他变量。这里是你的代码的问题：

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 

所以请尝试以下修正：

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));