Thread: base address of a simple array

Could any one tell how to get the base address of an array? or better if there is a predefined function that does this work?
Thanks in advance.

Here is what i did.

Code:

#include <stdio.h>

int main(void){
	int array[2][3]={1,2,3,4,5,6};
	int baseAddress = array; //should I write &array or *array ?? please help
	printf("base address of array is: %ld\n",baseAddress);
	return 0;
}

I have this but i have warnings...
"warning: initialization makes integer from pointer without a cast"

the result is always the same :

Code:

base address of array is: -1073743324

Use the %p format specifier:

Code:

#include <stdio.h>

int main(void){
	int array[2][3]={{1,2,3},{4,5,6}};
	printf("base address of array is: %p\n", (void*)array);
	return 0;
}

The base address of an array is always the name of the array. Therefore, in your example:

Code:

int array[...]

"array" is the base address. To printf out the address, do this:

Code:

printf("the address of 'array' is %p\n", array) ;

To look at the address of each array element, do this:

Code:

#include <stdio.h>
int main(int argc, char* argv[])
{
	int array[5] ; 

	int i ; 
	for (i = 0 ; i < sizeof(array) / sizeof(int) ; i++) 
	{ 
		printf("element %d is at address %p\n", i, &array[i]) ; 
	}
	return 0;
}

Todd

That is why i used a baseAddress variable.
How can i get the address without using %p ?

Code:

printf("the array is at address %x\n", array ) ;

To copy the address to another variable...

Code:

int * baseAddress = array ;

Originally Posted by Todd Burch

Code:

printf("the array is at address %x\n", array ) ;

That's however not guaranteed to work, since int and pointers are somtimes not the same size (64-bit machines for example have 64-bit pointers, whilst int are only 32-bit). %p is guaranteed to match the size of a pointer in architectures where pointers have a constant size - in architectures where that is not true, you may need modifiers or some other tricks - but fortunately non-constant size pointers are rare these days.

If you want to use an int to store a pointer, and subject to the pointer actually fitting in there:

Code:

int array[] = { ... };
int x = (int)array;

Using a cast tells the compiler "Look, I know it's not the right type, but just trust me, it's what I really want to do" - and of course, if that's not what you wanted, then you know who to blame [hint: it's not the compiler].

--
Mats

How can i get the address without using %p ?

If you want to print an address, any address, you must use %p and cast the pointer that holds that address to void*. If you want to store that address, then you store it in a pointer, in this case an int*.

thanks all

Originally Posted by brewbuck

The specific address is irrelevant, unless you are going to print it.

Actually I am working on Image processing program, and I am supposed to pass the base address of the bitmap to a function in order to get image represented in the screen.

I know it is not necessary an int.

The function is defined as follows

Code:

CGDataProviderRef CGDataProviderCreateWithData(
	void *info, //please ignore this
	cons void *data, //bitmap base address
	size_t size, //number of bytes
	CGDataProviderReleaseCallBack releaseDate //please ignore this
);

and supposing i have my bitmap as

Code:

char bitmapRGB[16*3][16*3] = {...};

I asked myself how could I get the base address of my bitmap?


by the way this is not the standard C ... is Objective-C for Mac OS but this function has no Objective-C grammar so is like C.

I asked myself how could I get the base address of my bitmap?

The bitmap base address is a const void* data, so you can just pass the array directly and it will be converted into a pointer to its first element. In C, pointers are implicitly cast to a pointer to void.

There's a little article that elaborate a little on pointers: http://cpwiki.sf.net/A_pointer_on_pointers
Remember: an address to something in the virtual memory is a pointer. So if you want to take the starting address of the array, what do you need? A pointer!