[Pgubook-readers] solution ch. 4 power.s

[Nick Gantt]

try adding a third call to the power function and add
it's result back in:

#PURPOSE: Program to illustrate how functions work
#         This program will compute the value of
#         2^3 + 5^2 + 4^2
#
#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 2nd argument
pushl $2        #push 1st argument
call power      #call the function
addl $8, %esp   #move the stack point back
pushl %eax      #save the first answer before
                #calling the next function
pushl $2        #push the 2nd argument
pushl $5        #push first argument
call power      #call the function
addl $8, %esp   #move stack pointer back
push %eax

push $2         #push 2nd argument              
push $4         #push 1st argument
call power      #call the function
addl $8, %esp   #move stack pointer back

popl %ebx       #The third answer is already in %eax.  We
                #saved the second answer onto the stack,
                #so now we can just pop it out into %ebx

addl %eax, %ebx #add 2nd and 3rd together, the result
is in %ebx

popl %ecx       #pop the first answer into %ecx
addl %ecx, %ebx #add 1st to sum of 2nd and 3rd, store
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 base 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 %ebx
movl 12(%ebp), %ecx     #put second arugment 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 to %eax
imull %ebx, %eax        #multiply the current result by the
base number
movl %eax, -4(%ebp)     #store 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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