Are you sure you compiled that particular piece of code with -S for generating asm output.........
gcc -S -o test.s test.c
my output shows that 20 bytes are subtracted, as shown :
test.c
void function(int a, int b, int c) {
char buff1[5];
char buff2[10];
}
void main() {
function(1,2,3);
}
Here's the assembly output :
------------------------------------------------------------------
.file "test.c"
.version "01.01"
gcc2_compiled.:
.text
.align 4
.globl function
.type function,@function
function:
pushl %ebp
movl %esp,%ebp
subl $20,%esp ************** Here it is !!! 20 bytes it is.
***********************************************************************
.L1:
leave
ret
.Lfe1:
.size function,.Lfe1-function
.align 4
.globl main
.type main,@function
main:
pushl %ebp
movl %esp,%ebp
pushl $3
pushl $2
pushl $1
call funciton
addl $12,%esp
.L2:
leave
ret
.Lfe2:
.size main,.Lfe2-main
.ident "GCC: (GNU) egcs-2.91.66 19990314/Linux (egcs-1.1.2 release)"
--------------------------------------------------------------------
Ali Saifullah Khan,
Project Administrator,
ConnPROBE Intrusion Detection System,
Sourceforge OpenSource Developer Networks.
>Am Mit, 2002-07-17 um 17.46 schrieb Jeremy Junginger:
>> In "Smashing the Stack for Fun and Profit" by Aleph One, There is a nice
>> example program called example1.c. It looks like this:
>>
>> void function(int a, int b, int c) {
>> char buffer1[5];
>> char buffer2[10];
>>
>> void main() {
>> function(1,2,3);
>> }
>>
>> Then, we go through how to generate assembley code output, how the
>> values are pushed onto the stack in reverse order, then the function
>> call, then moves the Frame Pointer onto the stack and copies the current
>> Stack Pointer into EBP. That part is groovy. Then when we look at the
>> function, in the example, he discusses how memory buffers are allocated
>> in "word" (4 byte) increments. That makes sense; however, when I
>> generate the assembly code with the exact same code, I see that it is
>> subtracting 40 rather than the expected 20
>> (bufger1(5bytes=2words=8bytes+10bytes=3words=12bytes). This part looks
>> crucial to understanding the rest of the concepts in the paper, so I'm
>> hesitant to continue without understanding this descrepancy. Any input
>> would be very much appreciated.
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This archive was generated by hypermail 2b30 : Wed Jul 17 2002 - 18:04:04 PDT