2006-09-02 Marcus Brinkmann <marcus@g10code.de>
[gnupg.git] / gl / allocsa.c
bloba54e2d0ca2f8e257bf161b9cd90220e77c0be1c0
1 /* Safe automatic memory allocation.
2 Copyright (C) 2003 Free Software Foundation, Inc.
3 Written by Bruno Haible <bruno@clisp.org>, 2003.
5 This program is free software; you can redistribute it and/or modify
6 it under the terms of the GNU General Public License as published by
7 the Free Software Foundation; either version 2, or (at your option)
8 any later version.
10 This program is distributed in the hope that it will be useful,
11 but WITHOUT ANY WARRANTY; without even the implied warranty of
12 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
13 GNU General Public License for more details.
15 You should have received a copy of the GNU General Public License
16 along with this program; if not, write to the Free Software Foundation,
17 Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */
19 #ifdef HAVE_CONFIG_H
20 # include <config.h>
21 #endif
23 /* Specification. */
24 #include "allocsa.h"
26 /* The speed critical point in this file is freesa() applied to an alloca()
27 result: it must be fast, to match the speed of alloca(). The speed of
28 mallocsa() and freesa() in the other case are not critical, because they
29 are only invoked for big memory sizes. */
31 #if HAVE_ALLOCA
33 /* Store the mallocsa() results in a hash table. This is needed to reliably
34 distinguish a mallocsa() result and an alloca() result.
36 Although it is possible that the same pointer is returned by alloca() and
37 by mallocsa() at different times in the same application, it does not lead
38 to a bug in freesa(), because:
39 - Before a pointer returned by alloca() can point into malloc()ed memory,
40 the function must return, and once this has happened the programmer must
41 not call freesa() on it anyway.
42 - Before a pointer returned by mallocsa() can point into the stack, it
43 must be freed. The only function that can free it is freesa(), and
44 when freesa() frees it, it also removes it from the hash table. */
46 #define MAGIC_NUMBER 0x1415fb4a
47 #define MAGIC_SIZE sizeof (int)
48 /* This is how the header info would look like without any alignment
49 considerations. */
50 struct preliminary_header { void *next; char room[MAGIC_SIZE]; };
51 /* But the header's size must be a multiple of sa_alignment_max. */
52 #define HEADER_SIZE \
53 (((sizeof (struct preliminary_header) + sa_alignment_max - 1) / sa_alignment_max) * sa_alignment_max)
54 struct header { void *next; char room[HEADER_SIZE - sizeof (struct preliminary_header) + MAGIC_SIZE]; };
55 /* Verify that HEADER_SIZE == sizeof (struct header). */
56 typedef int verify1[2 * (HEADER_SIZE == sizeof (struct header)) - 1];
57 /* We make the hash table quite big, so that during lookups the probability
58 of empty hash buckets is quite high. There is no need to make the hash
59 table resizable, because when the hash table gets filled so much that the
60 lookup becomes slow, it means that the application has memory leaks. */
61 #define HASH_TABLE_SIZE 257
62 static void * mallocsa_results[HASH_TABLE_SIZE];
64 #endif
66 void *
67 mallocsa (size_t n)
69 #if HAVE_ALLOCA
70 /* Allocate one more word, that serves as an indicator for malloc()ed
71 memory, so that freesa() of an alloca() result is fast. */
72 size_t nplus = n + HEADER_SIZE;
74 if (nplus >= n)
76 char *p = (char *) malloc (nplus);
78 if (p != NULL)
80 size_t slot;
82 p += HEADER_SIZE;
84 /* Put a magic number into the indicator word. */
85 ((int *) p)[-1] = MAGIC_NUMBER;
87 /* Enter p into the hash table. */
88 slot = (unsigned long) p % HASH_TABLE_SIZE;
89 ((struct header *) (p - HEADER_SIZE))->next = mallocsa_results[slot];
90 mallocsa_results[slot] = p;
92 return p;
95 /* Out of memory. */
96 return NULL;
97 #else
98 # if !MALLOC_0_IS_NONNULL
99 if (n == 0)
100 n = 1;
101 # endif
102 return malloc (n);
103 #endif
106 #if HAVE_ALLOCA
107 void
108 freesa (void *p)
110 /* mallocsa() may have returned NULL. */
111 if (p != NULL)
113 /* Attempt to quickly distinguish the mallocsa() result - which has
114 a magic indicator word - and the alloca() result - which has an
115 uninitialized indicator word. It is for this test that sa_increment
116 additional bytes are allocated in the alloca() case. */
117 if (((int *) p)[-1] == MAGIC_NUMBER)
119 /* Looks like a mallocsa() result. To see whether it really is one,
120 perform a lookup in the hash table. */
121 size_t slot = (unsigned long) p % HASH_TABLE_SIZE;
122 void **chain = &mallocsa_results[slot];
123 for (; *chain != NULL;)
125 if (*chain == p)
127 /* Found it. Remove it from the hash table and free it. */
128 char *p_begin = (char *) p - HEADER_SIZE;
129 *chain = ((struct header *) p_begin)->next;
130 free (p_begin);
131 return;
133 chain = &((struct header *) ((char *) *chain - HEADER_SIZE))->next;
136 /* At this point, we know it was not a mallocsa() result. */
139 #endif