assembly function flow

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:

1. what movl %ebx, -4(%ebp) #store current result for.
2. and what does "current result" means

in ma开发者_如何学Crked section in the code below, 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)

#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

Many assembly opcodes accept only one memory operand (either source, or destination). This probably explains why a move from memory to memory is done through %ebx.

I believe that this:

 movl 8(%ebp), %ebx #put first argument in %eax  

was a typo, and it should really be:

 movl 8(%ebp), %ebx #put first argument in %ebx  

and if you noticed, later the code is correct:

 movl %ebx, -4(%ebp) #store current result

In the end, the author could have used %eax for this operation as well (instead of %ebx), there's no reason why he shouldn't since it wouldn't change the program at all.

But the comment could be a lot clearer and I believe that this is a typo as well. At this point, it would be better if it said: #storing 1st argument on the local stack frame.

label power_loop_start uses that variable and temporarily stores it in %eax for quick operations and then place it back on the same location on the stack for the next loop:

 movl %eax, -4(%ebp)   #store the current result
 decl %ecx             #decrease the power
 jmp  power_loop_start #run for the next power

As Greg hinted, x86, like most mainstream architectures, does not have an instruction that copies data from memory to memory[1]. Thus, you must copy data using a separate load and store. First you load the data from the source memory into a register, then you store the data from that register to the destination memory. That's all that is happening here.

[1] I know, I know, but let's leave rep movs out of this and keep things simple.