static bool _cjose_jwe_set_iv_aes_cbc(cjose_jwe_t *jwe, cjose_err *err)
{
    // make sure we have an enc header
    json_t *enc_obj = json_object_get(jwe->hdr, CJOSE_HDR_ENC);
    if (NULL == enc_obj)
    {
        CJOSE_ERROR(err, CJOSE_ERR_INVALID_ARG);
        return false;
    }
    cjose_get_dealloc()(jwe->enc_iv.raw);

    // RFC 7516 (https://tools.ietf.org/html/rfc7516) is unclear about 
    // the size of the IV for AES-CBC. In section 5.1 
    // (https://tools.ietf.org/html/rfc7516#section-5.1), they state in no. 9.: 
    // "Generate a random JWE Initialization Vector of the correct size
    // for the content encryption algorithm"
    // And in the example in A.2.4 (https://tools.ietf.org/html/rfc7516#appendix-A.2.4)
    // they provide an example for AES128-CBC, which results (naturally) in the IV size of 128Bit. 
    //
    // The CISCO implementation chooses for the size of the IV the key size of the 
    // cipher algorithm, which seems to be wrong.
    //
    // According to RFC 3602 section 3 (https://tools.ietf.org/html/rfc3602#section-3): 
    // "The IV field MUST be the same size as the block size of the cipher algorithm being used."
    // And because the block size for AES cipher is always 16 Byte, the IV must be 16 Byte long.
    //
    // IV size for AES CBC is always 16 Byte
    jwe->enc_iv.raw_len = 16;

    // generate IV as random iv_size * 8 bit value
    if (!_cjose_jwe_malloc(jwe->enc_iv.raw_len, true, &jwe->enc_iv.raw, err))
    {
        return false;
    }

    return true;
}