how2heap分析系列:1_first_fit

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 int main()
 6 {
 7 printf("This file doesn‘t demonstrate an attack, but shows the nature of glibc‘s allocator.\n");
 8 printf("glibc uses a first-fit algorithm to select a free chunk.\n");
 9 printf("If a chunk is free and large enough, malloc will select this chunk.\n");
10 printf("This can be exploited in a use-after-free situation.\n");
11 
12 printf("Allocating 2 buffers. They can be large, don‘t have to be fastbin.\n");
13 char* a = malloc(512);
14 char* b = malloc(256);
15 char* c;
16 
17 printf("1st malloc(512): %p\n", a);
18 printf("2nd malloc(256): %p\n", b);
19 printf("we could continue mallocing here...\n");
20 printf("now let‘s put a string at a that we can read later \"this is A!\"\n");
21 strcpy(a, "this is A!");
22 printf("first allocation %p points to %s\n", a, a);
23 
24 printf("Freeing the first one...\n");
25 free(a);
26 
27 printf("We don‘t need to free anything again. As long as we allocate less than 512, it will end up at %p\n", a);
28 
29 printf("So, let‘s allocate 500 bytes\n");
30 c = malloc(500);
31 printf("3rd malloc(500): %p\n", c);
32 printf("And put a different string here, \"this is C!\"\n");
33 strcpy(c, "this is C!");
34 printf("3rd allocation %p points to %s\n", c, c);
35 printf("first allocation %p points to %s\n", a, a);
36 printf("If we reuse the first allocation, it now holds the data from the third allocation.");
37 }

junmoxiao@ubuntu:~/pwn/how2heap$ ./first_fit 
This file doesn‘t demonstrate an attack, but shows the nature of glibc‘s allocator.
glibc uses a first-fit algorithm to select a free chunk.
If a chunk is free and large enough, malloc will select this chunk.
This can be exploited in a use-after-free situation.
Allocating 2 buffers. They can be large, don‘t have to be fastbin.
1st malloc(512): 0x245b420
2nd malloc(256): 0x245b630
we could continue mallocing here...
now let‘s put a string at a that we can read later "this is A!"
first allocation 0x245b420 points to this is A!
Freeing the first one...
We don‘t need to free anything again. As long as we allocate less than 512, it will end up at 0x245b420
So, let‘s allocate 500 bytes
3rd malloc(500): 0x245b420
And put a different string here, "this is C!"
3rd allocation 0x245b420 points to this is C!
first allocation 0x245b420 points to this is C!
If we reuse the first allocation, it now holds the data from the third allocation.