将void指针转换为整数数组

Question

I have a problem where I have a pointer to an area in memory. I would like to use this pointer to create an integer array.

Essentially this is what I have, a pointer to a memory address of size 100*300*2 = 60000 bytes

unsigned char *ptr = 0x00000000; // fictional point in memory goes up to 0x0000EA60

What i would like to achieve is to examine this memory as an integer array of size 100*150 = 15000 ints = 60000 bytes, like this:

unsigned int array[ 100 ][ 150 ];

I'm assuming it involves some casting though i'm not sure exactly how to formulate it. Any help would be appreciated.

Answer 1

You can cast the pointer to unsigned int (*)[150] . It can then be used as if it is a 2D array ("as if", since behavior of sizeof is different).

unsigned int (*array)[150] = (unsigned int (*)[150]) ptr;

Answer 2

Starting with your ptr declaration

unsigned char *ptr = 0x00000000; // fictional point in memory goes up to 0x0000EA60

You can cast ptr to a pointer to whatever type you're treating the block as, in this case array of array of unsigned int. We'll declare a new pointer:

unsigned int (*array_2d)[100][150] = (unsigned int (*)[100][150])ptr;

Then, access elements by dereferencing and then indexing just as you would for a normal 2d array.

(*array_2d)[50][73] = 27;

Some typedef s would help clean things up, too.

typedef unsigned int my_2d_array_t[100][150];
typedef my_2d_array_t *my_2d_array_ptr_t;
my_2d_array_ptr_t array_2d = (my_2d_array_ptr_t)ptr;
(*array_2d)[26][3] = 357;
...

And sizeof should work properly.

sizeof(array_2d); //4, given 32-bit pointer
sizeof(*array_2d); //60000, given 32-bit ints
sizeof((*array_2d)[0]); //600, size of array of 150 ints
sizeof((*array_2d)[0][1]); //4, size of 1 int