CHR License Structure

[bimaindrawans]

I used the license.py file from your previous repository update, but I encountered an issue when generating a license for the CHR version. Would you be willing to share the structure for creating a license for the CHR type? This is because the software ID structure is different from the regular x86, which only has 9 characters, while in CHR there is a combination of letters and symbols totaling 11 characters.

should i cut the chr license byte like this ?

def generate_license(software_id, private_key: bytes, version: int = 7, level: int = 6):
    assert isinstance(private_key, bytes)
    
    if isinstance(software_id, str):
        # Hash the software_id if it's longer than 6 bytes (for CHR version)
        if len(software_id) > 6:
            hashed_id = hashlib.sha256(software_id.encode()).digest()
            software_id = int.from_bytes(hashed_id[:6], 'little')
        else:
            software_id = mikro_softwareid_decode(software_id)
    
    lic = software_id.to_bytes(6, 'little')
    lic += version.to_bytes(1, 'little')
    lic += level.to_bytes(1, 'little')
    lic += b'\0' * 8
    
    sig = mikro_kcdsa_sign(lic, private_key)
    lic = mikro_base64_encode(mikro_encode(lic) + sig, True)
    
    return (MIKRO_LICENSE_HEADER + '\n' + 
            lic[:len(lic) // 2] + '\n' + 
            lic[len(lic) // 2:] + '\n' + 
            MIKRO_LICENSE_FOOTER)

[israelrmm]

you must check this

https://t.me/mikrotikpatch/13486

[bimaindrawans]

you must check this

https://t.me/mikrotikpatch/13486

i didn't find chr structure on that script, because i think the structure written on the license_generator files.

[israelrmm]

you must check this

https://t.me/mikrotikpatch/13486

if you see that source code. RouterOS Lisecnse Key XXXX-XXXX has 6bit, and "little" is true....

chr lisense could be 3bit and "little" is false .. (See attchment)

then maybe lic = software_id.to_bytes(6, 'little') ---> lic = software_id.to_bytes(3, ???) ???=big?

and in "micro.py" ( mikro_softwareid_decode() function ) SOFTWARE_ID_CHARACTER_TABLE = b'TN0BYX18S5HZ4IA67DGF3LPCJQRUK9MW2VE' must be --->

SOFTWARE_ID_CHARACTER_TABLE = ???????????????????? <- This varible for chr is the trick and we need to found it.

(elsif = chyvalrys = zyb) apparently they are the same person... He decode it ( Revert ingeniering chr character table ) we mus be capable to do too...

[vnxme]

if you see that source code. RouterOS Lisecnse Key XXXX-XXXX has 6bit, and "little" is true....

chr lisense could be 3bit and "little" is false .. (See attchment)

then maybe lic = software_id.to_bytes(6, 'little') ---> lic = software_id.to_bytes(3, ???) ???=big?

No, 6 and 3 have nothing to do with the number of bytes a software ID can take in the license payload. These numbers indicate a license level. For CHR, the third level corresponds to the P-Unlimited.

[israelrmm]

if you see that source code. RouterOS Lisecnse Key XXXX-XXXX has 6bit, and "little" is true....
chr lisense could be 3bit and "little" is false .. (See attchment)
then maybe lic = software_id.to_bytes(6, 'little') ---> lic = software_id.to_bytes(3, ???) ???=big?

No, 6 and 3 have nothing to do with the number of bytes a software ID can take in the license payload. These numbers indicate a license level. For CHR, the third level corresponds to the P-Unlimited.

Ok. i understand. Thanks