Thread: Any way to dynamically change an array size in C?

Hello

Is there any way to dynamically change the size of an array?

For example -

Code:

unsigned char result[] = {0x01, 0x01}

// do stuff with result

// now I need to resize result so it can store three bytes, e.g. {0x01, 0x01, 0x01}

Can this be done?

Thanks!

No. (You may want to look into the exciting world of malloc() and free().)

Thank you for the replies.

Do you mean this kind of thing?

Code:

unsigned char *result;

// I want my array to store two bytes
result = (unsigned char *) malloc(2);

// Assign values to both array elements
*(result+0) = 0x01;
*(result+1) = 0xFF;

free(result);

// Now I want my array to store three bytes
result = (unsigned char *) malloc(3);

// Assign values to the three array elements
*(result+0) = 0x01;
*(result+1) = 0xFF;
*(result+1) = 0xCA;

free(result);

Thanks again.

Right. If you're often going to keep the original parts the same, like you do here, then you can look into realloc().

And also, you are free to use array notation on pointers too: result[2] instead of *(result+2).

The key here is realloc().

Edit: Beaten by tabstop.

best to use sizeof() for malloc.

[code]

result = malloc(2 * sizeof(char));

/*assign */

result = realloc(result, 3 * sizeof(char)); /*can be intinsive for large memory blocks */

/*then call free()*/

[code/]

Thanks everyone.

slingerland3g:

When I execute the following:

Code:

	unsigned char *result;

	result = malloc(2 * sizeof(unsigned char));

	printf("size of result: %d\n", sizeof *result);

I get the following:

size of result: 1

Surely this should be 2?

So it's size will always be 1 (according to sizeof), but I can still use result to store 2 unsigned chars?

Yes. One character (which is what *result is) is always one. sizeof does not report the entire memory block, just what you hand it.

Thanks everyone.

Good point King Mir, I was just being explicit and showing clear intent within the code.