سؤال

I am currently busy with assembly and ran into the following problem:

I'm trying to get a number that has been typed into the eax register. First i present a string which asks for input and then someone has to enter a number.

I've used the following code, but I don't understand every bit of it. Please note the comments in the code.

I know absolutely nothing happens now with the number, except that is has been moved into eax. What i do want to know is why i must use leal: why and what does it do? and why do i need to push eax back on the stack?

.text
string1: .asciz "Please enter a number\n"
input: .asciz "%d" 

.global main
main:
       # Editor's note: this code is broken / unsafe; missing push %ebp here
  movl %esp, %ebp
  
  push $string1          # prompt string
  call printf            #print the string
           # no add $4, %esp here: 4 bytes still allocated on the stack

  leal -4(%ebp), %eax   # ????
  pushl %eax            # the thing you pushed in eax is now pushed on the stack?
  pushl $input          #the number 
  
  call scanf      
  
  popl %eax
  popl %eax       # the number that has been entered is now in eax
  
  call end
  
end:
  push $0
  call exit
هل كانت مفيدة؟

المحلول

You are calling functions, so you pass parameters to them on the stack. One integer is returned to you in eax, rest is via input-output pointer parameters, again, on the stack. Check out x86 calling conventions.

Edit 0:

The leal instruction stores effective address of some temporary variable (that's where scanf puts your integer value) into eax, then you pass it to scanf on the stack. Take a look here: What's the purpose of the LEA instruction?

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