Showing posts with label sparc. Show all posts
Showing posts with label sparc. Show all posts

Monday, August 07, 2006

Stack Overflows in SPARC

Causing and Thereby Avoiding Sparc stack overflows
--------------------------------------------------

Have been hitting my head against the screen to get an
understanding of the overflow problem on a sparc machine.
Ran into some pretty good examples but it took me some
time to get a good grip on the way to do this.

for the 32bit version of things
===============================
sample program from http://blogs.sun.com/peteh

main()
call x();
x()
call y();
y()
int a[1];
a[20]=0


now why would this corrupt the stack and

which stack and
what part of the stack?


Let me try to answer these questions the best I can

PSEUDO-CODE ASSEMBLY
main() save %sp,... %fp=%sp
call x(); call x %o7=pc
x() save %sp,... %fp=%sp
call y(); call y %o7=pc
y() save %sp,... %fp=%sp
int a[1];
a[20]=0


definition int a[1];
What makes this offset 20 special?
Look at the .s file it should show you an offset of the a[] array
in mine it was showing these two lines for a[20]=0

add %fp,-20,%l0
st %g0,[%l0+80]


Thus it shows that a is pointing to fp-20


HigherAddr

fp of x() ----------------------
fr_argx[1] _____________________________
fr_argd[6] |nothing else coz there are no|
fr_stret |local variables etc in x() |
frame.h fr_savpc -----------------------------
from fr_savfp this is the registerw of main()
fr_arg[6]
fr_local[8]
fp of y() ----------------------
fp-20 a <---array a
register window of x
is saved here
current sp ----------------------

LowerAddr

so at fp you have fr_local %l0-%l7
at fp+8 you have fr_arg %i0-i5
at fp+14 you have fr_savfp %i6
at fp+15 you have fr_savpc %i7 <-----return addr

so the return address offset from a[] is
= +(20B+(15*4B))
= +60B
= +(20*4B)

so offset a[20] messes up the i7 for main and thereby the fault


now for the 64bit version of things
===================================
definition long a[1];
the assembly code shows

or %g0,1,%l3
add %fp,1991,%l1
stx %l3,[%l1+176]

so the array a is at fp+1991. Note the v9 arch specifies a 0x7ff stack bias
so this is within the stack even though you are adding to fp.

So where exactly is %o7 ?
Well 2047-1995 = 52B from the fp

fp+2047 -------------------

fp+1995 a


fp -------------------

the return address offset from a[] is

= +(52B+(15*8B))
= +172B
= +(22*8B)

so offset a[22] should corrupt i7 for main and cause a fault


So here we have managed to overwrite the %i7 register which stores
the return address. We have corrupted the stack of the x() function
so it cannot return correctly.




Sidenote: save instruction
Notice that most of the save instructions put in 96 bytes for the
stack. Why is that?

current sp ---------------------------------
%i0-%i7 and %l0-%l7

(input and local vars)


sp+64 --------------------------------- <- why 64? *A
hidden parameter
---------------------------------
%o0-%o5

(store first 6 params just in
case the params are passed by
addr and not by value)
sp+92 ---------------------------------
padding to be aligned
sp+96 --------------------------------- <- why pad? *B


*A - why 64? - 8*2 registers = 8*2*(8bytes ie 4 locations) = 64
*B - why pad> - this is SPARC land
refer this link http://www.cs.unm.edu/%7Emaccabe/classes/341/labman/node12.html
thats all!

Technorati Tags: , ,