I am looking at a presentation from MIT where they explain different types of ASLR implementations.
For example, they point out that for static ASLR, stack has 19-bits of entropy. In my understanding, this means the stack base address can only be randomized to take 2^19 different values.
I want to ask how to calculate that the stack has 19-bits of entropy ?
Edit:
After checking online, I found some explanation of stack ASLR on Linux. Learning from another question, the code I thought may be relevant is:
#ifndef STACK_RND_MASK
#define STACK_RND_MASK (0x7ff >> (PAGE_SHIFT - 12)) /* 8MB of VA */
#endif
static unsigned long randomize_stack_top(unsigned long stack_top)
{
unsigned int random_variable = 0;
if ((current->flags & PF_RANDOMIZE) &&
!(current->personality & ADDR_NO_RANDOMIZE)) {
random_variable = get_random_int() & STACK_RND_MASK;
random_variable <<= PAGE_SHIFT;
}
#ifdef CONFIG_STACK_GROWSUP
return PAGE_ALIGN(stack_top) + random_variable;
#else
return PAGE_ALIGN(stack_top) - random_variable;
#endif
}
I want to ask if this is the right place to reason about my question ?
Firstly, pages have to be aligned to 4096-byte boundaries, which effectively zeroes the lower 12 bits.
Next, the kernel splits the address space in to 0x00000000 - 0xbfffffff
for user memory and 0xc0000000 - 0xffffffff
for kernel memory. We only care about user memory here, so the kernel memory can be ignored.
If we further split up the user address space into three ranges:
Range | 2 MSBs
--------------------+--------
00000000 - 3fffffff | 00
40000000 - 7fffffff | 01
80000000 - bfffffff | 10
Generally we don't want the stack in the first range, since that's where the heap lives. This means we only have two possible combinations for the 2 most significant bits, 01
and 10
, effectively turning 2 bits into 1.
We have 32 bits in a 32-bit address, so we can get our 19-bits of stack entropy with:32 - (page alignment bits) - (memory partitioning bits) = 32 - 12 - 1 = 19
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