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 _______________________________________________ Pgubook-readers mailing list [email protected] http://lists.nongnu.org/mailman/listinfo/pgubook-readers
