Asembly program with mov function

Question

Can someone explain to me what this program does and how things change in it

.data
x: .long 15
.text
.global main
main:
movl $0, %eax
movl %eax, x
movl x, %ebx
mov $1, %eax
mov $0, %ebx
int $0x80

when I run it in the terminal it gives this

eax            0x5655618d          1448436109
ecx            0x33def597          870249879
edx            0xffffd100          -12032
ebx            0xf7fa4000          -134594560
esp            0xffffd0dc          0xffffd0dc
ebp            0xf7ffd020          0xf7ffd020 <_rtld_global>
esi            0xffffd194          -11884
edi            0xf7ffcb80          -134231168
eip            0x5655618d          0x5655618d <main>
eflags         0x246               [ PF ZF IF ]
cs             0x23                35
ss             0x2b                43
ds             0x2b                43
es             0x2b                43
fs             0x0                 0
gs             0x63                99

I don't rly understand how values change when using the mov function.

Which change to what? You've only shown one snapshot of register state, presumably before the first `movl $0, %eax` in `main` sets EAX = 0. In GDB, use `layout reg`, or `display /x $eax` ; `display /x $ebx` ; `display /x (int)x`, then do `stepi` a few times. ("Display" sets up that expression to print on every step, like how the `layout reg` window updates itself.) See the bottom of https://stackoverflow.com/tags/x86/info for GDB asm tips. — Peter Cordes, Dec 01 '22 at 10:59

score 0 · Answer 1 · answered Dec 08 '22 at 18:06

mov copies the value from the register or memory address on the left to the register or memory address on the right. For example:

movl $5,%eax
movl %eax,%ebx

After this sequence of instructions, both %eax and %ebx contain 5.

For a labeled address like x in your example, since you have 15 at that address, that's the value that you'll load from there:

.data
x: .long 15
.text
.global main
main:
movl x,%eax   // %eax now contains 15
addl $5,%eax  // add 5 to the value of %eax, it now contains 20
movl %eax,x   // store the value of %eax into the address pointed to by x.
              // x now contains 20, the 15 that used to be there is now gone.

mov $1, %eax
mov $0, %ebx
int $0x80     // exit

Asembly program with mov function

1 Answers1