Here is the code I have:
17 mov ebx,msg
18 mov edx,5
19 push ebx
I am debugging with gdb and here is the output I have:
Breakpoint 1, print () at hello.asm:17
(gdb) info register sp
sp: 0xbffff37c
(gdb) info stack
#0 print () at hello.asm:17
(gdb) step
(gdb) info register sp
sp: 0xbffff37c
(gdb) info stack
#0 print () at hello.asm:18
(gdb) step
(gdb) info register sp
sp: 0xbffff378
(gdb) info stack
#0 print () at hello.asm:19
Well obviously
push ebx
'decrements' the register sp by 4 bytes.. However when I type
info stack
I am still seeing print() at hello.asm:19
My question is, what is info stack showing me, and what is info register sp showing me? What is the relation between the stack register and the info stack?
when a new function is called, a new stack frame is set up. Each stack frame represents a function. Within that stack frame, when you push variables onto the stack, the stack pointer changes as the stack grows due to the push. The print() hello.asm:19 is gdb using the instruction pointer to show you where in your source execution is. You're in the print() function, line 19, which is "push %ebx" After you execute a callq to call another function, bt will show you the current stack frame has changed.
C-x a C-x 2 crl-2
in gdb will split your terminal into 3. Commands in the bottom pain, disassembly in the middle pane, and registers in the top pane. This way you can step and know where you're up to and what in the register state changes and when - which is really useful.
<rant> Until it crashes because gdb's ncurses interfaces (aka tui) is broken, the gdb gatekeepers don't care, don't accept patches that fix these crashes. If it gets to you too much you need to use a gdb front end, eg eclipse or insight which can show you the same information and may not crash due to stupid ncurses code</rant>