memcpy for multidimensional array

Question

Is there a way we can copy every element from one multidimensional array to another multidimensional array by just doing one memcpy operation?

int array1[4][4][4][4];
int array2[4][4][4][4];
int main()
{
    memset(&array1,1,sizeof(array1));
    memset(&array2,0,sizeof(array2));
    printf_all("value in array2 %d \n",array2[1][1][1][1]);
    memcpy(&array2,&array1,sizeof(array2));
    printf("memcopied in array2 from array1 \n");
    printf("value in array2 %d \n",array2[1][1][1][1]); //not printing 1
}

Mark Lakata · Accepted Answer · 2014-10-04 00:33:02Z

3

Your code is correct. You should not expect the output to show you a value of 1. You should expect it to show you a value of 16843009, assuming a 4 byte int.

The reason is: you are filling array1 with bytes of value 1, not with ints of value 1. i.e. binary 00000001000000010000000100000001 (0x01010101) is being filled into all the int elements with your memset operation.

So regardless of the size of int on your machine (unless it's a single byte!) you should not expect to see the value 1.

I hope this helps.

edited Oct 4, 2014 at 0:33

Mark Lakata

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answered Oct 4, 2014 at 0:11

Darren Stone

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Comments

Johannes · Accepted Answer · 2014-10-03 23:38:59Z

2

Yes, your code should already be correct.

You have to consider memory layout when doing this. The arrays are all in one block, multi dimensional is essentially a math trick done by the compiler.

Your code says copy this memory content to the other memory block. since both share the same layout they will contain the same values.

The following code also just copies the values, but access is handled differently so you would have to think about how to get the order of elements correct.

int array1[4][4][4][4]; //elements 256
int array2[256]; 

int main()
{
    memcpy(&array2,&array1,sizeof(array1)); //will also copy
    // original access via: a + 4 * b + 16 * c + 64 * d
}

edited Oct 3, 2014 at 23:38

answered Oct 3, 2014 at 23:29

Johannes

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2 Comments

Kerrek SB Over a year ago

Why do you have to consider anything? array1 and array2 have the same type, so it should be possible to assign values by copying the underlying representation.

Johannes Over a year ago

Suppose you wanted to copy something different. Like an array of char * or a big char buffer where the elements are seperated by \0. Saying it like this it seems obvious but that is from a real world example.

AnT stands with Russia · Accepted Answer · 2014-10-04 00:42:29Z

Multidimensional array in C is a flat block of memory with no internal structure. Memory layout of a multidimensional array is exactly the same as that of a 1-dimensional array of the same total size. The multidimensional interface is implemented through simple index recalculation. You can always memcpy the whole multidimensional array exactly as you do it in your code.

This, of course, only applies to built-in multidimensional arrays, explicitly declared as such (as in your code sample). If you implement a hand-made multidimensional array as an array of pointers to sub-arrays, that data structure will not be copyable in one shot with memcpy.

However, apparently you have some misconceptions about how memset works. Your memset(&array1,1,sizeof(array1)); will not fill the array with 1s, meaning that your code is not supposed to print 1 regardless of which array you print. memset interprets target memory as an array of chars, not as an array of ints.

memset can be used to set memory to zero. As for non-zero values, memset is generally unsuitable for initializing arrays of any type other than char.

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