A slight typo here probably. Printing i1 would always result in the same address regardless of the double free.
7.1 KiB
Double Free
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Basic Information
If you free a block of memory more than once, it can mess up the allocator's data and open the door to attacks. Here's how it happens: when you free a block of memory, it goes back into a list of free chunks (e.g. the "fast bin"). If you free the same block twice in a row, the allocator detects this and throws an error. But if you free another chunk in between, the double-free check is bypassed, causing corruption.
Now, when you ask for new memory (using malloc
), the allocator might give you a block that's been freed twice. This can lead to two different pointers pointing to the same memory location. If an attacker controls one of those pointers, they can change the contents of that memory, which can cause security issues or even allow them to execute code.
Example:
#include <stdio.h>
#include <stdlib.h>
int main() {
// Allocate memory for three chunks
char *a = (char *)malloc(10);
char *b = (char *)malloc(10);
char *c = (char *)malloc(10);
char *d = (char *)malloc(10);
char *e = (char *)malloc(10);
char *f = (char *)malloc(10);
char *g = (char *)malloc(10);
char *h = (char *)malloc(10);
char *i = (char *)malloc(10);
// Print initial memory addresses
printf("Initial allocations:\n");
printf("a: %p\n", (void *)a);
printf("b: %p\n", (void *)b);
printf("c: %p\n", (void *)c);
printf("d: %p\n", (void *)d);
printf("e: %p\n", (void *)e);
printf("f: %p\n", (void *)f);
printf("g: %p\n", (void *)g);
printf("h: %p\n", (void *)h);
printf("i: %p\n", (void *)i);
// Fill tcache
free(a);
free(b);
free(c);
free(d);
free(e);
free(f);
free(g);
// Introduce double-free vulnerability in fast bin
free(h);
free(i);
free(h);
// Reallocate memory and print the addresses
char *a1 = (char *)malloc(10);
char *b1 = (char *)malloc(10);
char *c1 = (char *)malloc(10);
char *d1 = (char *)malloc(10);
char *e1 = (char *)malloc(10);
char *f1 = (char *)malloc(10);
char *g1 = (char *)malloc(10);
char *h1 = (char *)malloc(10);
char *i1 = (char *)malloc(10);
char *i2 = (char *)malloc(10);
// Print initial memory addresses
printf("After reallocations:\n");
printf("a1: %p\n", (void *)a1);
printf("b1: %p\n", (void *)b1);
printf("c1: %p\n", (void *)c1);
printf("d1: %p\n", (void *)d1);
printf("e1: %p\n", (void *)e1);
printf("f1: %p\n", (void *)f1);
printf("g1: %p\n", (void *)g1);
printf("h1: %p\n", (void *)h1);
printf("i1: %p\n", (void *)i1);
printf("i2: %p\n", (void *)i2);
return 0;
}
In this example, after filling the tcache with several freed chunks (7), the code frees chunk h
, then chunk i
, and then h
again, causing a double free (also known as Fast Bin dup). This opens the possibility of receiving overlapping memory addresses when reallocating, meaning two or more pointers can point to the same memory location. Manipulating data through one pointer can then affect the other, creating a critical security risk and potential for exploitation.
Executing it, note how i1
and i2
got the same address:
Initial allocations:
a: 0xaaab0f0c22a0
b: 0xaaab0f0c22c0
c: 0xaaab0f0c22e0
d: 0xaaab0f0c2300
e: 0xaaab0f0c2320
f: 0xaaab0f0c2340
g: 0xaaab0f0c2360
h: 0xaaab0f0c2380
i: 0xaaab0f0c23a0
After reallocations:
a1: 0xaaab0f0c2360
b1: 0xaaab0f0c2340
c1: 0xaaab0f0c2320
d1: 0xaaab0f0c2300
e1: 0xaaab0f0c22e0
f1: 0xaaab0f0c22c0
g1: 0xaaab0f0c22a0
h1: 0xaaab0f0c2380
i1: 0xaaab0f0c23a0
i2: 0xaaab0f0c23a0
Examples
- Dragon Army. Hack The Box
- We can only allocate Fast-Bin-sized chunks except for size
0x70
, which prevents the usual__malloc_hook
overwrite. - Instead, we use PIE addresses that start with
0x56
as a target for Fast Bin dup (1/2 chance). - One place where PIE addresses are stored is in
main_arena
, which is inside Glibc and near__malloc_hook
- We target a specific offset of
main_arena
to allocate a chunk there and continue allocating chunks until reaching__malloc_hook
to get code execution.
- We can only allocate Fast-Bin-sized chunks except for size
- zero_to_hero. PicoCTF
- Using Tcache bins and a null-byte overflow, we can achieve a double-free situation:
- We allocate three chunks of size
0x110
(A
,B
,C
) - We free
B
- We free
A
and allocate again to use the null-byte overflow - Now
B
's size field is0x100
, instead of0x111
, so we can free it again - We have one Tcache-bin of size
0x110
and one of size0x100
that point to the same address. So we have a double free.
- We allocate three chunks of size
- We leverage the double free using Tcache poisoning
- Using Tcache bins and a null-byte overflow, we can achieve a double-free situation:
References
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- Join the 💬 Discord group or the telegram group or follow us on Twitter 🐦 @hacktricks_live.
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