Array of uint8_t arrays

Question

I have four uint8_t arrays:

uint8_t arrayOne[12]   = { 0x00,0x01,0x00,0x00,0x00,0x06,0xFE,0x03,0x01,0xC1,0x00,0x01 };

uint8_t arrayTwo[12]   = { 0x00,0x01,0x00,0x00,0x00,0x06,0xFE,0x03,0x4E,0x2D,0x00,0x0C };

uint8_t arrayThree[12] = { 0x00,0x01,0x00,0x00,0x00,0x06,0xFE,0x03,0x01,0xF3,0x00,0x01 };

uint8_t arrayFour[12]  = { 0x00,0x01,0x00,0x00,0x00,0x06,0xFE,0x03,0x20,0x04,0x00,0x01 };

and I want them to add to another array:

uint8_t theArray[4][12] = { arrayOne,arrayTwo,arrayThree,arrayFour };

but the values of arrays change when I add them to theArray.

Why? How could I add them to array properly?

Please specify what you want to do: copy them into a single 2d array, or create an array[4] of these array [pointers]. "Add" has too wide meaning here. (Both strategies already answered, btw.) — user3125367, Jun 03 '15 at 10:09

score 4 · Accepted Answer · answered Jun 03 '15 at 09:55

The toplevel array should just be an array of pointers:

uint8_t *theArrays[] = { arrayOne,arrayTwo,arrayThree,arrayFour };

You will lose information about the lengths of each "row", but that's fine.

I don't think you can reference an array in the initializer like in your example and have the elements copied into a larger array automatically.

score 3 · Answer 2 · answered Jun 03 '15 at 10:00

You have to copy each array in the result array using standard function memcpy declared in header <string.h>. For example

#include <string.h>
#include <stdint.h>

//...

uint8_t * arrayPtrs[] = { arrayOne, arrayTwo, arrayThree, arrayFour };

for ( size_t i = 0; i < sizeof( theArray ) / sizeof( *theArray ); i++ )
{
    memcpy( theArray[i], arrayPtrs[i], sizeof( theArray[i] ) );
}

score 1 · Answer 3 · answered Jun 03 '15 at 09:58

1

You are trying to initialize an uint8_t array with pointers to uint8_t, not uint8_t elements.

try this:

uint8_t *theArrays[] = { array1,array2, ... };

answered Jun 03 '15 at 09:58

Noam M

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Array of uint8_t arrays

3 Answers3