Read 16 bits from memory in Assembly x86 NASM

Question

I'm trying to do a very simple exercise in assembly: sum N numbers located in contiguous memory cells. This is my actual code:

global _start 
section .data

        array: DD 300,10,20,30,40,50    ; Allocate the numbers
        arrayLen: EQU $-array           ; Get the quantity of numbers 

section .text
        _start:

        mov ecx, 0                       ; The counter
        mov ebx, 0                       ; The Accumulator

loop:   add ebx, DWORD [array+ecx]        ; get the i-th word  and add it to the accumulator
        add ecx, 4
        cmp ecx, arrayLen
        jne loop

        mov eax, 1
        int 0x80

The program is compiled and executed correctly but returns an incorrect value. The result should be 450, instead I obtain 194.

I noticed that 194 is the the same bitstream as 450, but the 9th bit is omitted. Debugging my program I argued that for some reason, that I can't understand, when I read

[array+ecx]

It reads only 8 bits although I specified the keyword DWORD.

Can someone help me? Thanks in advance

Where do you see the 8-bit read? ebx is increased by a 4-byte (dword) value, that's all right. Maybe you were looking at BL, not EBX? — nullptr
When I execute the program on the terminal, it returns the value in ebx. To see this returned value, I type echo $? — sirnino

nullptr nullptr · Accepted Answer · 2015-01-11T14:20:31

The program sums up the array correctly. The problem is with returning the result.

[Answer corrected, thanks to Jester.]

You pass the return value to sys_exit() (this is what mov eax, 1; int 0x80 does). sys_exit() leaves only the lower 8 bits of the return value (other bits are used for some flags).

That's where the 9th bit is lost.

Shell observes already truncated return value and prints it out.

Read 16 bits from memory in Assembly x86 NASM

1 Answers