Fill an array with the numbers from 1 to 10

Ten slots of memory, filled with the numbers 1 to 10. The thing to watch here is not the loop, which you have seen, but what it leaves behind in the memory panel: eighty bytes that a program elsewhere would call an array of ten numbers, and that memory itself has no opinion about at all.

Type 402000 into the memory panel. Eighty bytes: 01 then seven zeroes, 02 then seven zeroes, on to 0A. Seven eighths of what the program wrote is zero, because a qword is eight bytes wide whether or not the number in it needs them.

That is what resq 10 asked for. resq reserves units and not bytes, so the 10 means ten qwords and the program gets eighty bytes. Writing resb 10 by mistake gives you ten bytes, the loop writes past the end of them on its second pass, and nothing warns you, because there is nothing there to warn: numbers is an address and the loop is arithmetic on it.

The 8 in [numbers + rcx*8] is the same number for the same reason, and it has to match. Change the array to resd 10, the store to mov [numbers + rcx*4], eax and the scale to 4, and the loop fills ten dwords correctly. Change only two of the three and it writes the right values into the wrong places.

A loop that put the same value in every slot would not need to be a loop at all:

    lea rdi, [numbers]
    mov rax, 7
    mov rcx, 10
    cld
    rep stosq

Five instructions no matter how long the array is. That cannot help here, because every slot gets a different number, but it is the shape to reach for when they do not.