Cast void pointer to integer array

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.

How do you initially acquire the the address in ptr? It could be dangerous to cast an arbitrary pointer to unsigned int * if the initial pointer was not aligned to sizeof int. If that happens, then at best your accesses to the memory will be slow due to reads across cache lines. At worst, it could cause a bus error and your program will halt (depending on your cpu architecture). — Chris Rice
– Chris Rice, Commented Jul 13, 2012 at 2:05
hi chris, just saw your comment. em the address is actually similar to the one in the example and i'm not aquiring it from anywhere - it is an area in memory which i know to be free - working in an embedded environment. i can see what you mean though as this did cause an error when i had it aligned to a character then tried to cast to int. the (temporary) solution i am using is to manually input the address again as above — conor
– conor, Commented Jul 18, 2012 at 20:29

nhahtdh · Accepted Answer · 2012-07-12 14:55:44Z

22

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;

edited Jul 12, 2012 at 14:55

answered Jul 12, 2012 at 14:45

nhahtdh

56.9k15 gold badges131 silver badges164 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

conor Over a year ago

thanks yeah thats what i need. could you elaborate further on how to cast the pointer?

nhahtdh Over a year ago

Just cast it normally i.e. unsigned int (*array)[150] = (unsigned int (*)[150]) ptr;

user2284570 Over a year ago

@nhahtdh : I tried this :(unsigned long)result.c[2] = 126and I got this compiler error lvalue required as left operand of assignment.resultis a structure and cis of type unsigned short *.

nhahtdh Over a year ago

@user2284570: What are you trying to do? From what I see, it is irrelevant to the question here. But what you probably want is ((unsigned long*) result.c)[2] = 126

podom · Accepted Answer · 2017-11-02 19:09:26Z

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 typedefs 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

Collectives™ on Stack Overflow

Cast void pointer to integer array

2 Answers 2

4 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related