Question

I’m on a 64bit platform, so all memory adrs are 8 bytes.

So to get an estimate of the memory usage of an array, should I add 8 bytes to the sizeof(DATATYPE) for each entry in the array.

Example:

short unsigned int *ary = new short unsigned int[1000000]; //length 1mio
//sizeof(short unsinged int) = 2bytes 
//sizeof(short unsinged int*) = 8 bytes

So does each entry take up 10bytes? and will my 1mio length array therefore use atleast 10megabytes?

thanks

Answer 1

No, you don’t get a pointer for each and every array index. You get a single pointer pointing to the array, which is a contiguous block of memory, which is why the address of any index can be calculated from the index itself plus the array address.

For example, if the variable a known by the memory location 0xffff0012 is set to 0x76543210, then they could be laid out in memory as:

            +-------------+ This is on the stack or global.
0xffff0012  |  0x76543210 |
            +-------------+

            +-------------+ This is on the heap (and may also
0x76543210  |  a[     0]  |   have some housekeeping information).
            +-------------+
0x76543212  |  a[     1]  |
            +-------------+
0x76543214  |  a[     2]  |
            +-------------+
0x76543216  |  a[     3]  |
            +-------------+
               :       :
            +-------------+
0x7672B68E  |  a[999999]  |
            +-------------+

and you can see that the address of index n is 0x76543210 + n * 2.

So you will actually have one 8-byte pointer and a million 2-byte shorts which, in your case, totals 2,000,008 bytes.

This is on top of any malloc housekeeping overhead which, like the pointer itself, is minuscule compared to your actual array.