Question

gcc (GCC) 4.7.0 c89

I am allocating memory using the following:

db_data_size = 32;
db->db_data[i]->name = malloc(db_data_size);

(gdb) p db_data_size
$24 = 32
(gdb) p sizeof(db->db_data[i]->name)
$25 = 8
(gdb) n
205   db->db_data[i]->email = malloc(db_data_size);
(gdb) p sizeof(db->db_data[i]->name)
$26 = 8

In the debugger I get 8 bytes instead of the 32 bytes I think should have been allocated.

My structure is:

struct data {
    int id;
    int set;
    char *name;
    char *email;
};

struct database {
    struct data **db_data;
    size_t database_rows;
    size_t database_data_size;
};

The only think I can think of is that a char* is 8 bytes, and that is what I am getting. However, in malloc I have explicity asked for 32 bytes.

Answer 1

sizeof(db->db_data[i]->name)

tells you the size of

char *name;

which is the size of a pointer (to char). It does not tell you the size of the allocated block; if you need to remember that, you must store it separately.

So 8 is the correct answer for a pointer on a 64-bit system.