1/* strrchr (str, ch) -- Return pointer to last occurrence of CH in STR. 2 For Intel 80x86, x>=3. 3 Copyright (C) 1994-2022 Free Software Foundation, Inc. 4 This file is part of the GNU C Library. 5 6 The GNU C Library is free software; you can redistribute it and/or 7 modify it under the terms of the GNU Lesser General Public 8 License as published by the Free Software Foundation; either 9 version 2.1 of the License, or (at your option) any later version. 10 11 The GNU C Library is distributed in the hope that it will be useful, 12 but WITHOUT ANY WARRANTY; without even the implied warranty of 13 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU 14 Lesser General Public License for more details. 15 16 You should have received a copy of the GNU Lesser General Public 17 License along with the GNU C Library; if not, see 18 <https://www.gnu.org/licenses/>. */ 19 20#include <sysdep.h> 21#include "asm-syntax.h" 22 23#define PARMS 4+8 /* space for 2 saved regs */ 24#define RTN PARMS 25#define STR RTN 26#define CHR STR+4 27 28 .text 29ENTRY (strrchr) 30 31 pushl %edi /* Save callee-safe registers used here. */ 32 cfi_adjust_cfa_offset (4) 33 cfi_rel_offset (edi, 0) 34 pushl %esi 35 cfi_adjust_cfa_offset (4) 36 37 xorl %eax, %eax 38 movl STR(%esp), %esi 39 cfi_rel_offset (esi, 0) 40 movl CHR(%esp), %ecx 41 42 /* At the moment %ecx contains C. What we need for the 43 algorithm is C in all bytes of the dword. Avoid 44 operations on 16 bit words because these require an 45 prefix byte (and one more cycle). */ 46 movb %cl, %ch /* now it is 0|0|c|c */ 47 movl %ecx, %edx 48 shll $16, %ecx /* now it is c|c|0|0 */ 49 movw %dx, %cx /* and finally c|c|c|c */ 50 51 /* Before we start with the main loop we process single bytes 52 until the source pointer is aligned. This has two reasons: 53 1. aligned 32-bit memory access is faster 54 and (more important) 55 2. we process in the main loop 32 bit in one step although 56 we don't know the end of the string. But accessing at 57 4-byte alignment guarantees that we never access illegal 58 memory if this would not also be done by the trivial 59 implementation (this is because all processor inherent 60 boundaries are multiples of 4. */ 61 62 testl $3, %esi /* correctly aligned ? */ 63 jz L(19) /* yes => begin loop */ 64 movb (%esi), %dl /* load byte in question (we need it twice) */ 65 cmpb %dl, %cl /* compare byte */ 66 jne L(11) /* target found => return */ 67 movl %esi, %eax /* remember pointer as possible result */ 68L(11): orb %dl, %dl /* is NUL? */ 69 jz L(2) /* yes => return NULL */ 70 incl %esi /* increment pointer */ 71 72 testl $3, %esi /* correctly aligned ? */ 73 jz L(19) /* yes => begin loop */ 74 movb (%esi), %dl /* load byte in question (we need it twice) */ 75 cmpb %dl, %cl /* compare byte */ 76 jne L(12) /* target found => return */ 77 movl %esi, %eax /* remember pointer as result */ 78L(12): orb %dl, %dl /* is NUL? */ 79 jz L(2) /* yes => return NULL */ 80 incl %esi /* increment pointer */ 81 82 testl $3, %esi /* correctly aligned ? */ 83 jz L(19) /* yes => begin loop */ 84 movb (%esi), %dl /* load byte in question (we need it twice) */ 85 cmpb %dl, %cl /* compare byte */ 86 jne L(13) /* target found => return */ 87 movl %esi, %eax /* remember pointer as result */ 88L(13): orb %dl, %dl /* is NUL? */ 89 jz L(2) /* yes => return NULL */ 90 incl %esi /* increment pointer */ 91 92 /* No we have reached alignment. */ 93 jmp L(19) /* begin loop */ 94 95 /* We exit the loop if adding MAGIC_BITS to LONGWORD fails to 96 change any of the hole bits of LONGWORD. 97 98 1) Is this safe? Will it catch all the zero bytes? 99 Suppose there is a byte with all zeros. Any carry bits 100 propagating from its left will fall into the hole at its 101 least significant bit and stop. Since there will be no 102 carry from its most significant bit, the LSB of the 103 byte to the left will be unchanged, and the zero will be 104 detected. 105 106 2) Is this worthwhile? Will it ignore everything except 107 zero bytes? Suppose every byte of LONGWORD has a bit set 108 somewhere. There will be a carry into bit 8. If bit 8 109 is set, this will carry into bit 16. If bit 8 is clear, 110 one of bits 9-15 must be set, so there will be a carry 111 into bit 16. Similarly, there will be a carry into bit 112 24. If one of bits 24-31 is set, there will be a carry 113 into bit 32 (=carry flag), so all of the hole bits will 114 be changed. 115 116 3) But wait! Aren't we looking for C, not zero? 117 Good point. So what we do is XOR LONGWORD with a longword, 118 each of whose bytes is C. This turns each byte that is C 119 into a zero. */ 120 121 /* Each round the main loop processes 16 bytes. */ 122 123 /* Jump to here when the character is detected. We chose this 124 way around because the character one is looking for is not 125 as frequent as the rest and taking a conditional jump is more 126 expensive than ignoring it. 127 128 Some more words to the code below: it might not be obvious why 129 we decrement the source pointer here. In the loop the pointer 130 is not pre-incremented and so it still points before the word 131 we are looking at. But you should take a look at the instruction 132 which gets executed before we get into the loop: `addl $16, %esi'. 133 This makes the following subs into adds. */ 134 135 /* These fill bytes make the main loop be correctly aligned. 136 We cannot use align because it is not the following instruction 137 which should be aligned. */ 138 .byte 0, 0 139#ifndef PROF 140 /* Profiling adds some code and so changes the alignment. */ 141 .byte 0 142#endif 143 144L(4): subl $4, %esi /* adjust pointer */ 145L(41): subl $4, %esi 146L(42): subl $4, %esi 147L(43): testl $0xff000000, %edx /* is highest byte == C? */ 148 jnz L(33) /* no => try other bytes */ 149 leal 15(%esi), %eax /* store address as result */ 150 jmp L(1) /* and start loop again */ 151 152L(3): subl $4, %esi /* adjust pointer */ 153L(31): subl $4, %esi 154L(32): subl $4, %esi 155L(33): testl $0xff0000, %edx /* is C in third byte? */ 156 jnz L(51) /* no => try other bytes */ 157 leal 14(%esi), %eax /* store address as result */ 158 jmp L(1) /* and start loop again */ 159 160L(51): 161 /* At this point we know that the byte is in one of the lower bytes. 162 We make a guess and correct it if necessary. This reduces the 163 number of necessary jumps. */ 164 leal 12(%esi), %eax /* guess address of lowest byte as result */ 165 testb %dh, %dh /* is guess correct? */ 166 jnz L(1) /* yes => start loop */ 167 leal 13(%esi), %eax /* correct guess to second byte */ 168 169L(1): addl $16, %esi /* increment pointer for full round */ 170 171L(19): movl (%esi), %edx /* get word (= 4 bytes) in question */ 172 movl $0xfefefeff, %edi /* magic value */ 173 addl %edx, %edi /* add the magic value to the word. We get 174 carry bits reported for each byte which 175 is *not* 0 */ 176 177 /* According to the algorithm we had to reverse the effect of the 178 XOR first and then test the overflow bits. But because the 179 following XOR would destroy the carry flag and it would (in a 180 representation with more than 32 bits) not alter then last 181 overflow, we can now test this condition. If no carry is signaled 182 no overflow must have occurred in the last byte => it was 0. */ 183 184 jnc L(20) /* found NUL => check last word */ 185 186 /* We are only interested in carry bits that change due to the 187 previous add, so remove original bits */ 188 xorl %edx, %edi /* (word+magic)^word */ 189 190 /* Now test for the other three overflow bits. */ 191 orl $0xfefefeff, %edi /* set all non-carry bits */ 192 incl %edi /* add 1: if one carry bit was *not* set 193 the addition will not result in 0. */ 194 195 /* If at least one byte of the word is C we don't get 0 in %edi. */ 196 jnz L(20) /* found NUL => check last word */ 197 198 /* Now we made sure the dword does not contain the character we are 199 looking for. But because we deal with strings we have to check 200 for the end of string before testing the next dword. */ 201 202 xorl %ecx, %edx /* XOR with word c|c|c|c => bytes of str == c 203 are now 0 */ 204 movl $0xfefefeff, %edi /* magic value */ 205 addl %edx, %edi /* add the magic value to the word. We get 206 carry bits reported for each byte which 207 is *not* 0 */ 208 jnc L(4) /* highest byte is C => examine dword */ 209 xorl %edx, %edi /* ((word^charmask)+magic)^(word^charmask) */ 210 orl $0xfefefeff, %edi /* set all non-carry bits */ 211 incl %edi /* add 1: if one carry bit was *not* set 212 the addition will not result in 0. */ 213 jnz L(3) /* C is detected in the word => examine it */ 214 215 movl 4(%esi), %edx /* get word (= 4 bytes) in question */ 216 movl $0xfefefeff, %edi /* magic value */ 217 addl %edx, %edi /* add the magic value to the word. We get 218 carry bits reported for each byte which 219 is *not* 0 */ 220 jnc L(21) /* found NUL => check last word */ 221 xorl %edx, %edi /* (word+magic)^word */ 222 orl $0xfefefeff, %edi /* set all non-carry bits */ 223 incl %edi /* add 1: if one carry bit was *not* set 224 the addition will not result in 0. */ 225 jnz L(21) /* found NUL => check last word */ 226 xorl %ecx, %edx /* XOR with word c|c|c|c => bytes of str == c 227 are now 0 */ 228 movl $0xfefefeff, %edi /* magic value */ 229 addl %edx, %edi /* add the magic value to the word. We get 230 carry bits reported for each byte which 231 is *not* 0 */ 232 jnc L(41) /* highest byte is C => examine dword */ 233 xorl %edx, %edi /* ((word^charmask)+magic)^(word^charmask) */ 234 orl $0xfefefeff, %edi /* set all non-carry bits */ 235 incl %edi /* add 1: if one carry bit was *not* set 236 the addition will not result in 0. */ 237 jnz L(31) /* C is detected in the word => examine it */ 238 239 movl 8(%esi), %edx /* get word (= 4 bytes) in question */ 240 movl $0xfefefeff, %edi /* magic value */ 241 addl %edx, %edi /* add the magic value to the word. We get 242 carry bits reported for each byte which 243 is *not* 0 */ 244 jnc L(22) /* found NUL => check last word */ 245 xorl %edx, %edi /* (word+magic)^word */ 246 orl $0xfefefeff, %edi /* set all non-carry bits */ 247 incl %edi /* add 1: if one carry bit was *not* set 248 the addition will not result in 0. */ 249 jnz L(22) /* found NUL => check last word */ 250 xorl %ecx, %edx /* XOR with word c|c|c|c => bytes of str == c 251 are now 0 */ 252 movl $0xfefefeff, %edi /* magic value */ 253 addl %edx, %edi /* add the magic value to the word. We get 254 carry bits reported for each byte which 255 is *not* 0 */ 256 jnc L(42) /* highest byte is C => examine dword */ 257 xorl %edx, %edi /* ((word^charmask)+magic)^(word^charmask) */ 258 orl $0xfefefeff, %edi /* set all non-carry bits */ 259 incl %edi /* add 1: if one carry bit was *not* set 260 the addition will not result in 0. */ 261 jnz L(32) /* C is detected in the word => examine it */ 262 263 movl 12(%esi), %edx /* get word (= 4 bytes) in question */ 264 movl $0xfefefeff, %edi /* magic value */ 265 addl %edx, %edi /* add the magic value to the word. We get 266 carry bits reported for each byte which 267 is *not* 0 */ 268 jnc L(23) /* found NUL => check last word */ 269 xorl %edx, %edi /* (word+magic)^word */ 270 orl $0xfefefeff, %edi /* set all non-carry bits */ 271 incl %edi /* add 1: if one carry bit was *not* set 272 the addition will not result in 0. */ 273 jnz L(23) /* found NUL => check last word */ 274 xorl %ecx, %edx /* XOR with word c|c|c|c => bytes of str == c 275 are now 0 */ 276 movl $0xfefefeff, %edi /* magic value */ 277 addl %edx, %edi /* add the magic value to the word. We get 278 carry bits reported for each byte which 279 is *not* 0 */ 280 jnc L(43) /* highest byte is C => examine dword */ 281 xorl %edx, %edi /* ((word^charmask)+magic)^(word^charmask) */ 282 orl $0xfefefeff, %edi /* set all non-carry bits */ 283 incl %edi /* add 1: if one carry bit was *not* set 284 the addition will not result in 0. */ 285 jz L(1) /* C is not detected => restart loop */ 286 jmp L(33) /* examine word */ 287 288L(23): addl $4, %esi /* adjust pointer */ 289L(22): addl $4, %esi 290L(21): addl $4, %esi 291 292 /* What remains to do is to test which byte the NUL char is and 293 whether the searched character appears in one of the bytes 294 before. A special case is that the searched byte maybe NUL. 295 In this case a pointer to the terminating NUL char has to be 296 returned. */ 297 298L(20): cmpb %cl, %dl /* is first byte == C? */ 299 jne L(24) /* no => skip */ 300 movl %esi, %eax /* store address as result */ 301L(24): testb %dl, %dl /* is first byte == NUL? */ 302 jz L(2) /* yes => return */ 303 304 cmpb %cl, %dh /* is second byte == C? */ 305 jne L(25) /* no => skip */ 306 leal 1(%esi), %eax /* store address as result */ 307L(25): testb %dh, %dh /* is second byte == NUL? */ 308 jz L(2) /* yes => return */ 309 310 shrl $16,%edx /* make upper bytes accessible */ 311 cmpb %cl, %dl /* is third byte == C */ 312 jne L(26) /* no => skip */ 313 leal 2(%esi), %eax /* store address as result */ 314L(26): testb %dl, %dl /* is third byte == NUL */ 315 jz L(2) /* yes => return */ 316 317 cmpb %cl, %dh /* is fourth byte == C */ 318 jne L(2) /* no => skip */ 319 leal 3(%esi), %eax /* store address as result */ 320 321L(2): popl %esi /* restore saved register content */ 322 cfi_adjust_cfa_offset (-4) 323 cfi_restore (esi) 324 popl %edi 325 cfi_adjust_cfa_offset (-4) 326 cfi_restore (edi) 327 328 ret 329END (strrchr) 330 331weak_alias (strrchr, rindex) 332libc_hidden_builtin_def (strrchr) 333