How to movl $0x41,%ecx

Question

I've managed 'Hello World' with Gnu as!

So, next thing is print 1 to 10 right? (Maybe in ruby)

At the moment, I'll be happy to print A closely followed by B. Here's what I have.

.section .text
  .globl _start

_start:
  # Print A
  movl  $4,%eax
  pushl $0x41
  movl  %esp,%ecx        # Would rather movl $0x41,%ecx
  movl  $1,%ebx
  movl  $1,%edx
  int   $0x80

  # Closely followed by B
  movl  $4,%eax
  incl  (%esp)           # Rather incl(%ecx) here
  movl  %esp,%ecx
  movl  $1,%ebx
  movl  $1,%edx
  int   $0x80

  movl  $1,%eax
  movl  $0,%ebx
  int   $0x80

And it actually works, but my question is, why can't I

  movl  $0x41,%ecx

To begin with, and then

  incl (%ecx)

a little later on?

It's not the same. It looks like `ecx` is being used as a base pointer of some sort. — Mysticial, Jul 17 '12 at 01:25
@Mysticial write is declared as: `ssize_t write(int fd, const void *buf, size_t count)`, ecx is buf — bluekeys, Jul 17 '12 at 01:42

score 0 · Accepted Answer · answered Jul 17 '12 at 03:05

0

For a sys_write, %ecx wants to point to where the character or characters reside in memory, not "be" the character to print. "incb" might be more correct than "incl" since you're only incrementing one byte - either "(%esp)" or "(%ecx)" ought to work, since they point to the same place at this point. Note that you're incrementing "contents" of memory, not the pointer to memory.

Best, Frank

answered Jul 17 '12 at 03:05

Frank Kotler

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Thanks Frank. That makes a lot of sense. – bluekeys Jul 17 '12 at 15:24

How to movl $0x41,%ecx

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