138 lines
5.2 KiB
Diff
138 lines
5.2 KiB
Diff
diff -u -r ../elfutils-0.159/libelf/elf_getarsym.c ./libelf/elf_getarsym.c
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--- ../elfutils-0.159/libelf/elf_getarsym.c 2014-05-18 16:32:15.000000000 +0200
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+++ ./libelf/elf_getarsym.c 2014-05-30 14:53:58.602211085 +0200
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@@ -45,6 +45,124 @@
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#include <dl-hash.h>
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#include "libelfP.h"
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+#ifdef __ANDROID__
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+/* Find the first occurrence of C in S. */
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+void *
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+rawmemchr (const void *s, int c_in)
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+{
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+ /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
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+ long instead of a 64-bit uintmax_t tends to give better
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+ performance. On 64-bit hardware, unsigned long is generally 64
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+ bits already. Change this typedef to experiment with
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+ performance. */
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+ typedef unsigned long int longword;
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+
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+ const unsigned char *char_ptr;
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+ const longword *longword_ptr;
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+ longword repeated_one;
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+ longword repeated_c;
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+ unsigned char c;
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+
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+ c = (unsigned char) c_in;
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+
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+ /* Handle the first few bytes by reading one byte at a time.
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+ Do this until CHAR_PTR is aligned on a longword boundary. */
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+ for (char_ptr = (const unsigned char *) s;
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+ (size_t) char_ptr % sizeof (longword) != 0;
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+ ++char_ptr)
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+ if (*char_ptr == c)
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+ return (void *) char_ptr;
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+
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+ longword_ptr = (const longword *) char_ptr;
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+
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+ /* All these elucidatory comments refer to 4-byte longwords,
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+ but the theory applies equally well to any size longwords. */
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+
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+ /* Compute auxiliary longword values:
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+ repeated_one is a value which has a 1 in every byte.
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+ repeated_c has c in every byte. */
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+ repeated_one = 0x01010101;
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+ repeated_c = c | (c << 8);
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+ repeated_c |= repeated_c << 16;
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+ if (0xffffffffU < (longword) -1)
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+ {
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+ repeated_one |= repeated_one << 31 << 1;
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+ repeated_c |= repeated_c << 31 << 1;
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+ if (8 < sizeof (longword))
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+ {
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+ size_t i;
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+
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+ for (i = 64; i < sizeof (longword) * 8; i *= 2)
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+ {
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+ repeated_one |= repeated_one << i;
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+ repeated_c |= repeated_c << i;
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+ }
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+ }
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+ }
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+
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+ /* Instead of the traditional loop which tests each byte, we will
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+ test a longword at a time. The tricky part is testing if *any of
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+ the four* bytes in the longword in question are equal to NUL or
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+ c. We first use an xor with repeated_c. This reduces the task
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+ to testing whether *any of the four* bytes in longword1 is zero.
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+
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+ We compute tmp =
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+ ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
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+ That is, we perform the following operations:
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+ 1. Subtract repeated_one.
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+ 2. & ~longword1.
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+ 3. & a mask consisting of 0x80 in every byte.
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+ Consider what happens in each byte:
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+ - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
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+ and step 3 transforms it into 0x80. A carry can also be propagated
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+ to more significant bytes.
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+ - If a byte of longword1 is nonzero, let its lowest 1 bit be at
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+ position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
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+ the byte ends in a single bit of value 0 and k bits of value 1.
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+ After step 2, the result is just k bits of value 1: 2^k - 1. After
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+ step 3, the result is 0. And no carry is produced.
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+ So, if longword1 has only non-zero bytes, tmp is zero.
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+ Whereas if longword1 has a zero byte, call j the position of the least
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+ significant zero byte. Then the result has a zero at positions 0, ...,
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+ j-1 and a 0x80 at position j. We cannot predict the result at the more
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+ significant bytes (positions j+1..3), but it does not matter since we
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+ already have a non-zero bit at position 8*j+7.
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+
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+ The test whether any byte in longword1 is zero is equivalent
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+ to testing whether tmp is nonzero.
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+
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+ This test can read beyond the end of a string, depending on where
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+ C_IN is encountered. However, this is considered safe since the
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+ initialization phase ensured that the read will be aligned,
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+ therefore, the read will not cross page boundaries and will not
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+ cause a fault. */
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+
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+ while (1)
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+ {
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+ longword longword1 = *longword_ptr ^ repeated_c;
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+
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+ if ((((longword1 - repeated_one) & ~longword1)
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+ & (repeated_one << 7)) != 0)
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+ break;
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+ longword_ptr++;
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+ }
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+
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+ char_ptr = (const unsigned char *) longword_ptr;
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+
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+ /* At this point, we know that one of the sizeof (longword) bytes
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+ starting at char_ptr is == c. On little-endian machines, we
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+ could determine the first such byte without any further memory
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+ accesses, just by looking at the tmp result from the last loop
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+ iteration. But this does not work on big-endian machines.
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+ Choose code that works in both cases. */
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+
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+ char_ptr = (unsigned char *) longword_ptr;
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+ while (*char_ptr != c)
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+ char_ptr++;
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+ return (void *) char_ptr;
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+}
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+#endif
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+
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static int
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read_number_entries (uint64_t *nump, Elf *elf, size_t *offp, bool index64_p)
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@@ -166,7 +284,7 @@
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/* We have an archive. The first word in there is the number of
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entries in the table. */
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- uint64_t n;
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+ uint64_t n = 0;
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size_t off = elf->start_offset + SARMAG + sizeof (struct ar_hdr);
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if (read_number_entries (&n, elf, &off, index64_p) < 0)
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{
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