Thread: Syntax Question for Array in C

Hello. I'm working on a project where I have to use someone's code. I know what this block of code does, I need help with the actual syntax. I'm confused on why the uint32_t is in parenthesis and how the code following that "&flash_array[x] works. Thanks in advance.

Code:

const uint32_t flash_block_address[NUM_OF_FLASH_BLOCKS] =
{
    (uint32_t)&flash_array[248 * 64], 	
    (uint32_t)&flash_array[249 * 64], 	
    (uint32_t)&flash_array[250 * 64], 	
    (uint32_t)&flash_array[251 * 64], 	
    (uint32_t)&flash_array[252 * 64], 	
    (uint32_t)&flash_array[253 * 64], 	
    (uint32_t)&flash_array[254 * 64], 	
    (uint32_t)&flash_array[255 * 64], 	
};

Whatever the type of flash_array is, (uint32_t) casts the address of the indexed element to another data type - I would assume a 32 bit unsigned integer.
The statement assigns the eight values to the array flash_block_address[NUM_OF_FLASH_BLOCKS].

uint32_t is part of C99 and is defined in stdint.h, it is indeed a 32 bit unsigned integer.

Take a look at these warnings I get in this sample usage of what you posted:

Code:

#include <stdio.h>
#include <stdint.h>

#define NUM_OF_FLASH_BLOCKS 8

char flash_array[300 * 64];

int main(void)
{

  const  uint32_t flash_block_address[NUM_OF_FLASH_BLOCKS] =
  {
    (uint32_t)&flash_array[248 * 64]    /*  <---warning */
  };

warning: cast from pointer to integer of different size
warning: initializer element is not computable at load time

A pointer is not a regular integer,
It is a specific sized integer determined by the system.
On my system it is 8 bytes. (64 bits)
If I were to chop off half of the bits to make it a uint32_t
it would certainly NOT be the same location in memory.
So just let the system decide what size a pointer is.

YOU WANT an array of pointers to the proper type.
Here I make it char * to match the type of flash_array[]:

Code:

  const  char * flash_block_address[NUM_OF_FLASH_BLOCKS] =
  {
    &flash_array[248 * 64],
    &flash_array[249 * 64]
  };

  flash_array[248 * 64] = 0x7d;

  printf("%02x \n", *flash_block_address[0] );

  return 0;
}

out:

Code:

>  gcc -Wall -W -pedantic casting_Q.c -o casting_Q.bin
 
>  ./casting_Q.bin
7d

So yours needs to be matched to the flash_array[] in your code in the same way.

If you must use an integer type for a pointer for some reason, there is a uintptr_t also defined in stdint.h that is the same size as a pointer.

Out of curiosity, what could you do with that address cast to integer?

Originally Posted by HowardL

Out of curiosity, what could you do with that address cast to integer?

I don't know, it's there never the less, and is guaranteed to be the size of a void* if you find a reason to do it.

Thanks for the responses, makes sense now. This code was for an embedded platform of which I do not know any details, so it's very hardware specific. I will be porting it to a different micro-controller.