CODEWITHSUNDEEP
assemblygccmov
Barzoius Mecca
Barzoius Mecca

Reputation: 3

Asembly program with mov function

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.

Upvotes: -1

Views: 42

Answers (1)

puppydrum64
puppydrum64

Reputation: 1688

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

Upvotes: 0

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