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</details>
## Sandbox loading process
<figure><imgsrc="../../../.gitbook/assets/image.png"alt=""><figcaption><p>Image from <ahref="http://newosxbook.com/files/HITSB.pdf">http://newosxbook.com/files/HITSB.pdf</a></p></figcaption></figure>
In the previous image it's possible to observe **how the sandbox will be loaded** when an application with the entitlement **`com.apple.security.app-sandbox`** is run.
The compiler will link `/usr/lib/libSystem.B.dylib` to the binary.
Then, **`libSystem.B`** will be calling other several functions until the **`xpc_pipe_routine`** sends the entitlements of the app to **`securityd`**. Securityd checks if the process should be quarantine inside the Sandbox, and if so, it will be quarentine.\
Finally, the sandbox will be activated will a call to **`__sandbox_ms`** which will call **`__mac_syscall`**.
### Sanbox load debug & bypass
Let's compile an application that should be sandboxed:
{% tabs %}
{% tab title="sand.c" %}
```c
#include <stdlib.h>
int main() {
system("cat ~/Desktop/del.txt");
}
```
{% endtab %}
{% tab title="entitlements.xml" %}
```xml
<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd"><plistversion="1.0">
[**This research**](https://saagarjha.com/blog/2020/05/20/mac-app-store-sandbox-escape/) discovered 2 ways to bypass the Sandbox. Because the sandbox is applied from userland when the **libSystem** library is loaded. If a binary could avoid loading it, it would never get sandboxed:
* If the binary was **completely statically compiled**, it could avoid loading that library.
* If the **binary wouldn't need to load any libraries** (because the linker is also in libSystem), it won't need to load libSystem. 
### Shellcodes
Note that **even shellcodes** in ARM64 needs to be linked in `libSystem.dylib`:
```bash
ld -o shell shell.o -macosx_version_min 13.0
ld: dynamic executables or dylibs must link with libSystem.dylib for architecture arm64
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