CODEWITHSUNDEEP
cassemblyx86
yiati
yiati

Reputation: 1025

x86 assembly code basic

What does the following code do...

   0x080489b0 <+0>: push   %ebp
   0x080489b1 <+1>: mov    %esp,%ebp
   0x080489b3 <+3>: sub    $0x14,%esp
   0x080489b6 <+6>: push   %ebx

To my knowledge this sets up the stack in the first two lines, and then pushes the %esp down 14 essentially increasing the stack on the third line, and finally pushes a %ebx onto the stack (even though it's empty), which also increases the stack, and pushes %esp down one more spot.

Am I wrong on any of these steps? Thanks for any help!

Upvotes: 0

Views: 723

Answers (2)

Adam Mihalcin
Adam Mihalcin

Reputation: 14478

You are roughly correct. The third instruction (sub $0x14,%esp) actually means to grow the stack by 0x14 = 20 bytes, perhaps for alignment reasons. The fourth line means to push the 4-byte contents of %ebx onto the stack, subtracting 4 from %esp as a side effect. The reason for this is that %ebx is a callee-save register: if a function modifies the contents of %ebx, it is expected to push %ebx onto the stack, then modify %ebx, then pop %ebx back off the stack so that the caller never notices a change in %ebx. The other callee-save registers on x86 are %esi and %edi.

Upvotes: 1

user149341
user149341

Reputation:

Not quite. sub $0x14, %esp subtracts hexadecimal 0x14 (decimal 20) from the stack pointer, which is equivalent to the space used by 5 dwords.

Upvotes: 1

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