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1 | diff -u -r ../elfutils-0.159/libelf/elf_getarsym.c ./libelf/elf_getarsym.c |
2 | --- ../elfutils-0.159/libelf/elf_getarsym.c 2014-05-18 16:32:15.000000000 +0200 | |
3 | +++ ./libelf/elf_getarsym.c 2014-05-30 14:53:58.602211085 +0200 | |
4 | @@ -45,6 +45,124 @@ | |
5 | #include <dl-hash.h> | |
6 | #include "libelfP.h" | |
7 | ||
8 | +#ifdef __ANDROID__ | |
9 | +/* Find the first occurrence of C in S. */ | |
10 | +void * | |
11 | +rawmemchr (const void *s, int c_in) | |
12 | +{ | |
13 | + /* On 32-bit hardware, choosing longword to be a 32-bit unsigned | |
14 | + long instead of a 64-bit uintmax_t tends to give better | |
15 | + performance. On 64-bit hardware, unsigned long is generally 64 | |
16 | + bits already. Change this typedef to experiment with | |
17 | + performance. */ | |
18 | + typedef unsigned long int longword; | |
19 | + | |
20 | + const unsigned char *char_ptr; | |
21 | + const longword *longword_ptr; | |
22 | + longword repeated_one; | |
23 | + longword repeated_c; | |
24 | + unsigned char c; | |
25 | + | |
26 | + c = (unsigned char) c_in; | |
27 | + | |
28 | + /* Handle the first few bytes by reading one byte at a time. | |
29 | + Do this until CHAR_PTR is aligned on a longword boundary. */ | |
30 | + for (char_ptr = (const unsigned char *) s; | |
31 | + (size_t) char_ptr % sizeof (longword) != 0; | |
32 | + ++char_ptr) | |
33 | + if (*char_ptr == c) | |
34 | + return (void *) char_ptr; | |
35 | + | |
36 | + longword_ptr = (const longword *) char_ptr; | |
37 | + | |
38 | + /* All these elucidatory comments refer to 4-byte longwords, | |
39 | + but the theory applies equally well to any size longwords. */ | |
40 | + | |
41 | + /* Compute auxiliary longword values: | |
42 | + repeated_one is a value which has a 1 in every byte. | |
43 | + repeated_c has c in every byte. */ | |
44 | + repeated_one = 0x01010101; | |
45 | + repeated_c = c | (c << 8); | |
46 | + repeated_c |= repeated_c << 16; | |
47 | + if (0xffffffffU < (longword) -1) | |
48 | + { | |
49 | + repeated_one |= repeated_one << 31 << 1; | |
50 | + repeated_c |= repeated_c << 31 << 1; | |
51 | + if (8 < sizeof (longword)) | |
52 | + { | |
53 | + size_t i; | |
54 | + | |
55 | + for (i = 64; i < sizeof (longword) * 8; i *= 2) | |
56 | + { | |
57 | + repeated_one |= repeated_one << i; | |
58 | + repeated_c |= repeated_c << i; | |
59 | + } | |
60 | + } | |
61 | + } | |
62 | + | |
63 | + /* Instead of the traditional loop which tests each byte, we will | |
64 | + test a longword at a time. The tricky part is testing if *any of | |
65 | + the four* bytes in the longword in question are equal to NUL or | |
66 | + c. We first use an xor with repeated_c. This reduces the task | |
67 | + to testing whether *any of the four* bytes in longword1 is zero. | |
68 | + | |
69 | + We compute tmp = | |
70 | + ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). | |
71 | + That is, we perform the following operations: | |
72 | + 1. Subtract repeated_one. | |
73 | + 2. & ~longword1. | |
74 | + 3. & a mask consisting of 0x80 in every byte. | |
75 | + Consider what happens in each byte: | |
76 | + - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, | |
77 | + and step 3 transforms it into 0x80. A carry can also be propagated | |
78 | + to more significant bytes. | |
79 | + - If a byte of longword1 is nonzero, let its lowest 1 bit be at | |
80 | + position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, | |
81 | + the byte ends in a single bit of value 0 and k bits of value 1. | |
82 | + After step 2, the result is just k bits of value 1: 2^k - 1. After | |
83 | + step 3, the result is 0. And no carry is produced. | |
84 | + So, if longword1 has only non-zero bytes, tmp is zero. | |
85 | + Whereas if longword1 has a zero byte, call j the position of the least | |
86 | + significant zero byte. Then the result has a zero at positions 0, ..., | |
87 | + j-1 and a 0x80 at position j. We cannot predict the result at the more | |
88 | + significant bytes (positions j+1..3), but it does not matter since we | |
89 | + already have a non-zero bit at position 8*j+7. | |
90 | + | |
91 | + The test whether any byte in longword1 is zero is equivalent | |
92 | + to testing whether tmp is nonzero. | |
93 | + | |
94 | + This test can read beyond the end of a string, depending on where | |
95 | + C_IN is encountered. However, this is considered safe since the | |
96 | + initialization phase ensured that the read will be aligned, | |
97 | + therefore, the read will not cross page boundaries and will not | |
98 | + cause a fault. */ | |
99 | + | |
100 | + while (1) | |
101 | + { | |
102 | + longword longword1 = *longword_ptr ^ repeated_c; | |
103 | + | |
104 | + if ((((longword1 - repeated_one) & ~longword1) | |
105 | + & (repeated_one << 7)) != 0) | |
106 | + break; | |
107 | + longword_ptr++; | |
108 | + } | |
109 | + | |
110 | + char_ptr = (const unsigned char *) longword_ptr; | |
111 | + | |
112 | + /* At this point, we know that one of the sizeof (longword) bytes | |
113 | + starting at char_ptr is == c. On little-endian machines, we | |
114 | + could determine the first such byte without any further memory | |
115 | + accesses, just by looking at the tmp result from the last loop | |
116 | + iteration. But this does not work on big-endian machines. | |
117 | + Choose code that works in both cases. */ | |
118 | + | |
119 | + char_ptr = (unsigned char *) longword_ptr; | |
120 | + while (*char_ptr != c) | |
121 | + char_ptr++; | |
122 | + return (void *) char_ptr; | |
123 | +} | |
124 | +#endif | |
125 | + | |
126 | ||
127 | static int | |
128 | read_number_entries (uint64_t *nump, Elf *elf, size_t *offp, bool index64_p) | |
129 | @@ -166,7 +284,7 @@ | |
130 | ||
131 | /* We have an archive. The first word in there is the number of | |
132 | entries in the table. */ | |
133 | - uint64_t n; | |
134 | + uint64_t n = 0; | |
135 | size_t off = elf->start_offset + SARMAG + sizeof (struct ar_hdr); | |
136 | if (read_number_entries (&n, elf, &off, index64_p) < 0) | |
137 | { |