通过引用传递并将值赋给struct的指针

Question

I have a struct MY_TYPE :

struct MY_TYPE {
    boolean flag;
    short int xyz;
};

I have another struct MY_DATA which has a field with type of pointer to the above struct:

struct MY_DATA {
    MY_TYPE *m_type;
    double value;
};

I have a function which takes a pointer to pointer of MY_DATA as parameter:

getData(struct MY_DATA **m_data) {
    // create a MY_TYPE pointer & assign value to its field
    struct MY_TYPE *m_type_tmp = malloc(sizeof(struct MY_TYPE));
    m_type_tmp -> xyz = 123;

    // I try to assign a value to the MY_TYPE field of m_data,
    // there is no compiler error, but run time error "bad access"
    (** m_data).m_type = m_type_tmp;
}

I call above function by:

struct MY_DATA *data;
get_data(&data);

The compiler doesn't complain anything, but when run my code, I get "Bad access" on the last line of code of function get_data(...) , how come?

Answer 1

Since you pass the address of an uninitialized pointer, you must allocate the structure MY_DATA as well:

void getData(struct MY_DATA **m_data) {
    // create a MY_TYPE pointer & assign value to its field
    struct MY_TYPE *m_type_tmp = malloc(sizeof(struct MY_TYPE));
    m_type_tmp->flag = 0;
    m_type_tmp->xyz = 123;

    // allocate the MY_DATA structure:
    *m_data = malloc(sizeof(*m_data));
    // Initialize all members
    (*m_data)->m_type = m_type_tmp;
    (*m_data)->value = 0;
}

It would be simpler to have the function return a pointer to the allocated structure:

struct MY_DATA *getData(void) {
    // create a MY_TYPE pointer & assign value to its field
    struct MY_TYPE *m_type_tmp = malloc(sizeof(*m_type_tmp));
    m_type_tmp->flag = 0;
    m_type_tmp->xyz = 123;

    // allocate the MY_DATA structure:
    struct MY_DATA *m_data = malloc(sizeof(*m_data));
    // Initialize all members
    m_data->m_type = m_type_tmp;
    m_data->value = 0;
    return m_data;
}

And invoke it this way:

struct MY_DATA *data = get_data();

Answer 2

getData(struct MY_DATA ** m_data)

You don't need pointer to pointer to MY_DATA , pointer to MY_DATA would be suffice.

Then instead of (** m_data).m_type = m_type_tmp; you can write:

m_data->m_type = m_type_tmp;

Also these are incorrect:

struct MY_DATA *data;  // <-- you haven't initialized the pointer
get_data(&data);

You can fix that by:

struct MY_DATA data = {0};
get_data(&data);

Answer 3

按引用传递意味着您将值传递给其他变量的引用，而按值传递意味着将值传递给字符串，float，character或boolean。