汇编函数流 [英] assembly function flow

问题描述

我读了从地上爬起来编程，如果你不知道这本书是什么，你仍然可以帮助我。

在这本书中（第4章）有2个东西，我不明白：

1. 什么 MOVL％EBX，-4（％EBP）#store当前结果为


2. 和什么是当前结果的意思

在下面的code标记部分，有：

  MOVL 8（EBP％），EBX％
 

这意味着保存 8（％EBP）到％EBX ，但之所以我不明白是，如果程序员希望 8（％EBP）保存到 -4（％EBP），何必 8（％EBP）到％EBX ？是 MOVL 8（EBP％）-4（％EBP）akward的？或者有没有在 MOVL 8（EBP％）的错字，EBX％在#put EAX％第一个参数？
（我认为％EBX 应％EAX 或反之亦然）

  #PURPOSE：程序来说明如何工作
＃这一方案将计算的价值
＃2 ^ 3 + 5 ^ 2
#Everything在主程序存储在寄存器
#so数据部分并没有什么。.section伪。数据
.section伪的.text
.globl _start_开始：pushl $ 3 #push第二个参数
pushl $ 2 #push第一个参数
调用功率#CALL功能
ADDL $ 8％ESP #move堆栈指针返回
pushl％eax中#save第一个答案之前#calling下一功能pushl $ 2 #push第二个参数
pushl $ 5 #push第一个参数调用功率#CALL功能
ADDL $ 8％ESP #move堆栈指针返回
popl％EBX＃系统第二个答案已经#in％eax中。我们保存了
#first回答压入堆栈，
#so现在我们只是弹出它
#out到％EBXADDL％EAX，EBX％＃将它们放在一起
＃系统结果是EBX％MOVL $ 1，％eax中#exit（返回％EBX）
INT 0x80的$#PURPOSE：该函数用于计算
＃数字的值升高到
＃电源。#INPUT：第一个参数 - 基数
＃第二个参数 - 电源
＃提高到
＃
#OUTPUT：会给结果作为返回值
＃
#NOTES：功率必须大于或等于1
＃
#VARIABLES：
＃％EBX  - 持有基数
＃％ECX  - 持有的力量
＃
＃-4（％EBP） - 保存当前结果
＃
＃％eax中用于临时存储
＃.TYPE功率，@function
功率：
pushl％ebp的#save老基指针
MOVL％ESP，EBP％了#make堆栈指针的基址指针
subl $ 4％ESP＃获取空间，我们的本地存储
##########################################MOVL 8（EBP％），EBX％在#put EAX％第一个参数
MOVL 12（％EBP），ECX％第二#put在ECX％论证
MOVL％EBX，-4（％EBP）#store当前结果##########################################power_loop_start：
CMPL $ 1，ECX％＃如果功率为1，我们已经完成
JE end_power
MOVL -4（％EBP），％eax中#move当前结果到EAX％
imull％EBX，EAX％的#multiply当前的结果通过＃系统基数
MOVL％EAX，-4（％EBP）#store当前结果
DECL％ECX #decrease电源
JMP power_loop_start #run下一个电源end_power：
MOVL -4（％EBP），％eax中#return价值远远在％eax中
MOVL％EBP，ESP％的#restore堆栈指针
popl％EBP #restore基指针
RET
 




解决方案

我相信这一点：

  MOVL 8（EBP％），EBX％在#put EAX％第一个参数
 

是一个错字，它确实应该：

  MOVL 8（EBP％），EBX％在#put％ebx中第一个参数
 

和你注意到没有，以后code是正确的：

  MOVL％EBX，-4（％EBP）#store当前结果
 

在最后，笔者也可以使用％EAX 此操作，以及（而不是％EBX ），没有理由他不应该，因为它不会改变程序的。

但是注释可以是更加清晰，我认为这是一个错字，以及。 本地堆栈帧上#storing第一个参数：在这一点上，如果说这将是更好的。

标签power_loop_start 使用这个变量，并暂时将其存储在％EAX 快速操作，然后将其放回在同一位置堆栈上下一个循环：

  MOVL％EAX，-4（％EBP）#store当前结果
 DECL％ECX #decrease电源
 JMP power_loop_start #run下一个电源
 

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 marked 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

解决方案

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

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