RT-AC66 3.0.0.4.374.130 core
[tomato.git] / release / src-rt-6.x / linux / linux-2.6 / fs / ecryptfs / debug.c
blob434c7efd80f814eea18662354d371fa9af0178ae
1 /**
2 * eCryptfs: Linux filesystem encryption layer
3 * Functions only useful for debugging.
5 * Copyright (C) 2006 International Business Machines Corp.
6 * Author(s): Michael A. Halcrow <mahalcro@us.ibm.com>
8 * This program is free software; you can redistribute it and/or
9 * modify it under the terms of the GNU General Public License as
10 * published by the Free Software Foundation; either version 2 of the
11 * License, or (at your option) any later version.
13 * This program is distributed in the hope that it will be useful, but
14 * WITHOUT ANY WARRANTY; without even the implied warranty of
15 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
16 * General Public License for more details.
18 * You should have received a copy of the GNU General Public License
19 * along with this program; if not, write to the Free Software
20 * Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA
21 * 02111-1307, USA.
24 #include "ecryptfs_kernel.h"
26 /**
27 * ecryptfs_dump_auth_tok - debug function to print auth toks
29 * This function will print the contents of an ecryptfs authentication
30 * token.
32 void ecryptfs_dump_auth_tok(struct ecryptfs_auth_tok *auth_tok)
34 char salt[ECRYPTFS_SALT_SIZE * 2 + 1];
35 char sig[ECRYPTFS_SIG_SIZE_HEX + 1];
37 ecryptfs_printk(KERN_DEBUG, "Auth tok at mem loc [%p]:\n",
38 auth_tok);
39 if (auth_tok->flags & ECRYPTFS_PRIVATE_KEY) {
40 ecryptfs_printk(KERN_DEBUG, " * private key type\n");
41 ecryptfs_printk(KERN_DEBUG, " * (NO PRIVATE KEY SUPPORT "
42 "IN ECRYPTFS VERSION 0.1)\n");
43 } else {
44 ecryptfs_printk(KERN_DEBUG, " * passphrase type\n");
45 ecryptfs_to_hex(salt, auth_tok->token.password.salt,
46 ECRYPTFS_SALT_SIZE);
47 salt[ECRYPTFS_SALT_SIZE * 2] = '\0';
48 ecryptfs_printk(KERN_DEBUG, " * salt = [%s]\n", salt);
49 if (auth_tok->token.password.flags &
50 ECRYPTFS_PERSISTENT_PASSWORD) {
51 ecryptfs_printk(KERN_DEBUG, " * persistent\n");
53 memcpy(sig, auth_tok->token.password.signature,
54 ECRYPTFS_SIG_SIZE_HEX);
55 sig[ECRYPTFS_SIG_SIZE_HEX] = '\0';
56 ecryptfs_printk(KERN_DEBUG, " * signature = [%s]\n", sig);
58 ecryptfs_printk(KERN_DEBUG, " * session_key.flags = [0x%x]\n",
59 auth_tok->session_key.flags);
60 if (auth_tok->session_key.flags
61 & ECRYPTFS_USERSPACE_SHOULD_TRY_TO_DECRYPT)
62 ecryptfs_printk(KERN_DEBUG,
63 " * Userspace decrypt request set\n");
64 if (auth_tok->session_key.flags
65 & ECRYPTFS_USERSPACE_SHOULD_TRY_TO_ENCRYPT)
66 ecryptfs_printk(KERN_DEBUG,
67 " * Userspace encrypt request set\n");
68 if (auth_tok->session_key.flags & ECRYPTFS_CONTAINS_DECRYPTED_KEY) {
69 ecryptfs_printk(KERN_DEBUG, " * Contains decrypted key\n");
70 ecryptfs_printk(KERN_DEBUG,
71 " * session_key.decrypted_key_size = [0x%x]\n",
72 auth_tok->session_key.decrypted_key_size);
73 ecryptfs_printk(KERN_DEBUG, " * Decrypted session key "
74 "dump:\n");
75 if (ecryptfs_verbosity > 0)
76 ecryptfs_dump_hex(auth_tok->session_key.decrypted_key,
77 ECRYPTFS_DEFAULT_KEY_BYTES);
79 if (auth_tok->session_key.flags & ECRYPTFS_CONTAINS_ENCRYPTED_KEY) {
80 ecryptfs_printk(KERN_DEBUG, " * Contains encrypted key\n");
81 ecryptfs_printk(KERN_DEBUG,
82 " * session_key.encrypted_key_size = [0x%x]\n",
83 auth_tok->session_key.encrypted_key_size);
84 ecryptfs_printk(KERN_DEBUG, " * Encrypted session key "
85 "dump:\n");
86 if (ecryptfs_verbosity > 0)
87 ecryptfs_dump_hex(auth_tok->session_key.encrypted_key,
88 auth_tok->session_key.
89 encrypted_key_size);
93 /**
94 * ecryptfs_dump_hex - debug hex printer
95 * @data: string of bytes to be printed
96 * @bytes: number of bytes to print
98 * Dump hexadecimal representation of char array
100 void ecryptfs_dump_hex(char *data, int bytes)
102 int i = 0;
103 int add_newline = 1;
105 if (ecryptfs_verbosity < 1)
106 return;
107 if (bytes != 0) {
108 printk(KERN_DEBUG "0x%.2x.", (unsigned char)data[i]);
109 i++;
111 while (i < bytes) {
112 printk("0x%.2x.", (unsigned char)data[i]);
113 i++;
114 if (i % 16 == 0) {
115 printk("\n");
116 add_newline = 0;
117 } else
118 add_newline = 1;
120 if (add_newline)
121 printk("\n");