On Sun, Mar 20, 2011 at 6:59 PM, 김태윤
<address@hidden> wrote:
assembly function flow
hello
I am reading a "programming from the ground up"
if you don't know what this book is, you still can help me.
in this book(chapter 4) there are 2 things that I don't understand.
Q. I don't understand
1. what "movl %ebx, -4(%ebp) #store current result" for.
2. and what does "current result" means
in marked section in the code below
little upperside, there is
"movl 8(%ebp), %ebx" which means save 8(%ebp) to %ebx
but the reason why I don't understand is
if the programmer want 8(%ebp) to save to -4(%ebp),
why should 8(%ebp) be passed through %ebx?
is "movl 8(%ebp), -4(%ebp)" akward?
or is there any typo in "movl 8(%ebp), %ebx #put first argument in %eax"?
(I think %ebx should be %eax or vice versa)
Thanks in advanced
#PURPOSE: Program to illustrate how functions work
# This program will compute the value of
# 2^3 + 5^2
#
#Everything in the main program is stored in registers,
#so the data section doesn’t have anything.
.section .data
.section .text
.globl _start
_start:
pushl $3 #push second argument
pushl $2 #push first argument
call power #call the function
addl $8, %esp #move the stack pointer back
pushl %eax #save the first answer before
#calling the next function
pushl $2 #push second argument
pushl $5 #push first argument
call power #call the function
addl $8, %esp #move the stack pointer back
popl %ebx #The second answer is already
#in %eax. We saved the
#first answer onto the stack,
#so now we can just pop it
#out into %ebx
addl %eax, %ebx #add them together
#the result is in %ebx
movl $1, %eax #exit (%ebx is returned)
int $0x80
#PURPOSE: This function is used to compute
# the value of a number raised to
# a power.
#
#INPUT: First argument - the base number
# Second argument - the power to
# raise it to
#
#OUTPUT: Will give the result as a return value
#
#NOTES: The power must be 1 or greater
#
#VARIABLES:
# %ebx - holds the base number
# %ecx - holds the power
#
# -4(%ebp) - holds the current result
#
# %eax is used for temporary storage
#
.type power, @function
power:
pushl %ebp #save old base pointer
movl %esp, %ebp #make stack pointer the base pointer
subl $4, %esp #get room for our local storage
##########################################
movl 8(%ebp), %ebx #put first argument in %eax
movl 12(%ebp), %ecx #put second argument in %ecx
movl %ebx, -4(%ebp) #store current result
##########################################
power_loop_start:
cmpl $1, %ecx #if the power is 1, we are done
je end_power
movl -4(%ebp), %eax #move the current result into %eax
imull %ebx, %eax #multiply the current result by
#the base number
movl %eax, -4(%ebp) #store the current result
decl %ecx #decrease the power
jmp power_loop_start #run for the next power
end_power:
movl -4(%ebp), %eax #return value goes in %eax
movl %ebp, %esp #restore the stack pointer
popl %ebp #restore the base pointer
ret
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