Basic assembly

Basic assembly

[theguest]

Hi!
First of all, I'm spanish so sorry for my bad english.
I'm new on the list and I'm a novice in assembly. I'm interested in
linux assembly because I'm interested in linux security, buffer
overflows...
So, I'm reading the "Smash the stack for fun and profit" and the results
that Aleph get are different from mine.
For example he has this code in C:
-------------
void function(int a, int b, int c){
	char buffer1[5];
	char buffer2[10];
}
void main(){
	function (1,2,3);
}
-------------
He gets linux assembly code using gcc:
$ gcc -S -o example1.s example1.c

He says his assembly code has a line like:
subl $20,%esp
He says the memory is reserved multiples of "word". He says word=4bytes.
So our 5 bytes buffer needs 2 words = 8 bytes and buffer2[10]->12bytes.
That's why uses subl to get 20 bytes from esp. That seems correct for
me but when I compile it in assembly using gcc I don't get the same
assembly code...
For example:
	subl $8,%esp
	addl $-4,%esp
in the main function 
and in the function <function>:			
	subl $40,%esp
Why are my code reserving other quantity of space for variables?
It's something related with optimizated compilation in gcc?
Thanks in advance and sorry if something isn't correct in my first email
to the list.

RE: Basic assembly

[Joseph D. Wagner]

> First of all, I'm spanish so sorry for my bad english.

Your English is better than some other people's English.

> He says the memory is reserved multiples of "word".

No.  In Linux, memory is allocated in "page frames", where 1 page frame
= 4 Kbytes.

> He says word=4bytes.

No.  A Word is 2 bytes.

WORD (Word) = 2 bytes (16 bits)
DWORD (Double Word) = 4 bytes (32 bits)
QWORD (Quad Word) = 8 bytes (64 bits)

> So our 5 bytes buffer needs = 8 bytes ...
> ...
> Why are my code reserving other quantity of space for variables?

All variables must be DWORD "aligned" except for variables smaller than
1 DWORD.

"DWORD Alignment" means that the amount of memory required for all
variables will be rounded up to the nearest DWORD (4 bytes).

For example, given the user defined class:
class Pentagon {
unsigned char Side1;
unsigned char Side2;
unsigned char Side3;
unsigned char Side4;
unsigned char Side5;
};
while the size of Pentagon is 5 bytes, the compiler rounds up allocation
to 8 bytes so that Pentagon is DWORD Aligned.

DWORD Alignment occurs with each instance of the variable.  In other
words, an array of Pentagon, for example: Pentagon myPentagon[5];, is 25
bytes but would round up to 40 bytes (not 28 bytes) because each
instance of Pentagon is 8 bytes, so an array of 5 Pentagons would be 40
bytes (8 byte alignment * 5 occurrences = 40 bytes).

DWORD Alignment has nothing to do with Linux, the compiler, or any
computer language.  DWORD Alignment is part of the x86 processor
specification.  (A DWORD is 32 bits, and today's processors are 32 bit
processors; that is not a coincidence.)

I am not sure if this helps you.  Your specific problem was unclear to
me.

By the way, whoever is giving you your information is really
misinformed.