CODEWITHSUNDEEP
assemblyx86
user1210446
user1210446

Reputation: 57

Buffer in x86 Assembly

Could someone help explain this code to me?

.text:00401270 ; int __cdecl main(int argc,const char **argv,const char *envp)
.text:00401270 Dst = byte ptr −80h

...More Code...

.text:00401270 push ebp
.text:00401271 mov ebp, esp
.text:00401273 sub esp, 80h
.text:00401293 push 80h
.text:00401298 push 0
.text:0040129A lea eax, [ebp+Dst]
.text:0040129D push eax
.text:0040129E call _memset

I get that a buffer of size 0x80 is created and filled with the value 0 when _memset is called. However I do not understand the usage of the pointer [ebp+Dst]. Why is the base pointer (ebp) involved at all? Additionally, why is Dst set to a negative value?

Upvotes: 1

Views: 2564

Answers (1)

jcomeau_ictx
jcomeau_ictx

Reputation: 38422

it's ebp, not edp; it is being used to access the stack where esp pointed before the 80-byte buffer is placed on it. then Dst, -80, is added, which points to the start (low byte) of the buffer. there is no need to do it this way in assembly, these constructs are the compiler's rendition of the C code.

Upvotes: 2

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