如何在结构数组中获取结构地址？

Question

I'm trying to get struct's address.

I want to get address in an int * , and I want to change address by adding numbers to the int * . I tried several ways, but I can't solve it.

struct num_d {
    unsigned char data;    
    unsigned char pad1;
    unsigned char pad2;
    unsigned char pad3;
};

struct num_d **m = malloc(sizeof(struct num_d *) * row);   

for (int i = 0; i < row; i++)                       
{                                                   
    m[i] = malloc(sizeof(struct num_d) * col);    
}

How can I get m[0][0] 's address in an int * ?

Answer 1

first things first lets typedef your struct, so we can type less and be more clear:

typedef struct num_d num_d;

void pointer

A pointer to void is a "generic" pointer type. A void * can be converted to any other pointer type without an explicit cast . we cannot de-reference a void * or do pointer arithmetic with it; you must convert it to a complete data type pointer first (like int* eg) then do the de-refrence or the pointer arithmetic.

Now, malloc() return a void* which points to the allocated heap buffer (if malloc successed in allocation other wise null is the return value).

you code become:

num_d** m = malloc(sizeof(num_d*) * row);  /*m is an array of void* pointers (not initialized)*/ 

for (int i = 0; i < row; i++)                       
{                                                   
    m[i] = malloc(sizeof(num_d) * col); /*in each element in m you have a void* that points to struct num_d on the heap*/    
}

the sizeof(void*) is the same as sizeof any pointer (except function pointers in some machines/os).

putting it all together

How can I get m[0][0]'s address in an int *?

This is a wrong question! because m is an array of void* to "num_d structs" (holding the num_d heap address).

if you want the start address of the i-th num_d struct in the array m , then, just return the void* in the index i in this array m[i] . and if you want to cast it just cast it (no need actually) just assign it:

int* ptr = m[i];

Take in mind that compilers will warn you, regarding the assignment above (but this assignment is supported and legal):

warning: initialization from incompatible pointer type [-Wincompatible-pointer-types]

or (no need again):

int* ptr = (int*)m[i];

I don't know why you need such behavior, it makes more sense to cast to num_d*

if you want the address of the first data member in the struct num_d , then you must cast to the appropriate data type to get the expected data:

unsigned char data = ((num_d*)m[i])->data;

unsigned char* p_data = &((num_d*)m[i])->data;

Answer 2

You don't need to have the address in an int* in order to be adding to it. The way that [] works, is that it adds to the pointer and dereferences.

You can just add to *(m[0] + 1) to get the second element.

Answer 3

How about:

int *ptr = (int *) m[0];