Source file src/cmd/compile/internal/ssa/magic.go
1 // Copyright 2016 The Go Authors. All rights reserved. 2 // Use of this source code is governed by a BSD-style 3 // license that can be found in the LICENSE file. 4 5 package ssa 6 7 import ( 8 "math/big" 9 "math/bits" 10 ) 11 12 // So you want to compute x / c for some constant c? 13 // Machine division instructions are slow, so we try to 14 // compute this division with a multiplication + a few 15 // other cheap instructions instead. 16 // (We assume here that c != 0, +/- 1, or +/- 2^i. Those 17 // cases are easy to handle in different ways). 18 19 // Technique from https://gmplib.org/~tege/divcnst-pldi94.pdf 20 21 // First consider unsigned division. 22 // Our strategy is to precompute 1/c then do 23 // ⎣x / c⎦ = ⎣x * (1/c)⎦. 24 // 1/c is less than 1, so we can't compute it directly in 25 // integer arithmetic. Let's instead compute 2^e/c 26 // for a value of e TBD (^ = exponentiation). Then 27 // ⎣x / c⎦ = ⎣x * (2^e/c) / 2^e⎦. 28 // Dividing by 2^e is easy. 2^e/c isn't an integer, unfortunately. 29 // So we must approximate it. Let's call its approximation m. 30 // We'll then compute 31 // ⎣x * m / 2^e⎦ 32 // Which we want to be equal to ⎣x / c⎦ for 0 <= x < 2^n-1 33 // where n is the word size. 34 // Setting x = c gives us c * m >= 2^e. 35 // We'll chose m = ⎡2^e/c⎤ to satisfy that equation. 36 // What remains is to choose e. 37 // Let m = 2^e/c + delta, 0 <= delta < 1 38 // ⎣x * (2^e/c + delta) / 2^e⎦ 39 // ⎣x / c + x * delta / 2^e⎦ 40 // We must have x * delta / 2^e < 1/c so that this 41 // additional term never rounds differently than ⎣x / c⎦ does. 42 // Rearranging, 43 // 2^e > x * delta * c 44 // x can be at most 2^n-1 and delta can be at most 1. 45 // So it is sufficient to have 2^e >= 2^n*c. 46 // So we'll choose e = n + s, with s = ⎡log2(c)⎤. 47 // 48 // An additional complication arises because m has n+1 bits in it. 49 // Hardware restricts us to n bit by n bit multiplies. 50 // We divide into 3 cases: 51 // 52 // Case 1: m is even. 53 // ⎣x / c⎦ = ⎣x * m / 2^(n+s)⎦ 54 // ⎣x / c⎦ = ⎣x * (m/2) / 2^(n+s-1)⎦ 55 // ⎣x / c⎦ = ⎣x * (m/2) / 2^n / 2^(s-1)⎦ 56 // ⎣x / c⎦ = ⎣⎣x * (m/2) / 2^n⎦ / 2^(s-1)⎦ 57 // multiply + shift 58 // 59 // Case 2: c is even. 60 // ⎣x / c⎦ = ⎣(x/2) / (c/2)⎦ 61 // ⎣x / c⎦ = ⎣⎣x/2⎦ / (c/2)⎦ 62 // This is just the original problem, with x' = ⎣x/2⎦, c' = c/2, n' = n-1. 63 // s' = s-1 64 // m' = ⎡2^(n'+s')/c'⎤ 65 // = ⎡2^(n+s-1)/c⎤ 66 // = ⎡m/2⎤ 67 // ⎣x / c⎦ = ⎣x' * m' / 2^(n'+s')⎦ 68 // ⎣x / c⎦ = ⎣⎣x/2⎦ * ⎡m/2⎤ / 2^(n+s-2)⎦ 69 // ⎣x / c⎦ = ⎣⎣⎣x/2⎦ * ⎡m/2⎤ / 2^n⎦ / 2^(s-2)⎦ 70 // shift + multiply + shift 71 // 72 // Case 3: everything else 73 // let k = m - 2^n. k fits in n bits. 74 // ⎣x / c⎦ = ⎣x * m / 2^(n+s)⎦ 75 // ⎣x / c⎦ = ⎣x * (2^n + k) / 2^(n+s)⎦ 76 // ⎣x / c⎦ = ⎣(x + x * k / 2^n) / 2^s⎦ 77 // ⎣x / c⎦ = ⎣(x + ⎣x * k / 2^n⎦) / 2^s⎦ 78 // ⎣x / c⎦ = ⎣(x + ⎣x * k / 2^n⎦) / 2^s⎦ 79 // ⎣x / c⎦ = ⎣⎣(x + ⎣x * k / 2^n⎦) / 2⎦ / 2^(s-1)⎦ 80 // multiply + avg + shift 81 // 82 // These can be implemented in hardware using: 83 // ⎣a * b / 2^n⎦ - aka high n bits of an n-bit by n-bit multiply. 84 // ⎣(a+b) / 2⎦ - aka "average" of two n-bit numbers. 85 // (Not just a regular add & shift because the intermediate result 86 // a+b has n+1 bits in it. Nevertheless, can be done 87 // in 2 instructions on x86.) 88 89 // umagicOK reports whether we should strength reduce a n-bit divide by c. 90 func umagicOK(n uint, c int64) bool { 91 // Convert from ConstX auxint values to the real uint64 constant they represent. 92 d := uint64(c) << (64 - n) >> (64 - n) 93 94 // Doesn't work for 0. 95 // Don't use for powers of 2. 96 return d&(d-1) != 0 97 } 98 99 // umagicOKn reports whether we should strength reduce an unsigned n-bit divide by c. 100 // We can strength reduce when c != 0 and c is not a power of two. 101 func umagicOK8(c int8) bool { return c&(c-1) != 0 } 102 func umagicOK16(c int16) bool { return c&(c-1) != 0 } 103 func umagicOK32(c int32) bool { return c&(c-1) != 0 } 104 func umagicOK64(c int64) bool { return c&(c-1) != 0 } 105 106 type umagicData struct { 107 s int64 // ⎡log2(c)⎤ 108 m uint64 // ⎡2^(n+s)/c⎤ - 2^n 109 } 110 111 // umagic computes the constants needed to strength reduce unsigned n-bit divides by the constant uint64(c). 112 // The return values satisfy for all 0 <= x < 2^n 113 // 114 // floor(x / uint64(c)) = x * (m + 2^n) >> (n+s) 115 func umagic(n uint, c int64) umagicData { 116 // Convert from ConstX auxint values to the real uint64 constant they represent. 117 d := uint64(c) << (64 - n) >> (64 - n) 118 119 C := new(big.Int).SetUint64(d) 120 s := C.BitLen() 121 M := big.NewInt(1) 122 M.Lsh(M, n+uint(s)) // 2^(n+s) 123 M.Add(M, C) // 2^(n+s)+c 124 M.Sub(M, big.NewInt(1)) // 2^(n+s)+c-1 125 M.Div(M, C) // ⎡2^(n+s)/c⎤ 126 if M.Bit(int(n)) != 1 { 127 panic("n+1st bit isn't set") 128 } 129 M.SetBit(M, int(n), 0) 130 m := M.Uint64() 131 return umagicData{s: int64(s), m: m} 132 } 133 134 func umagic8(c int8) umagicData { return umagic(8, int64(c)) } 135 func umagic16(c int16) umagicData { return umagic(16, int64(c)) } 136 func umagic32(c int32) umagicData { return umagic(32, int64(c)) } 137 func umagic64(c int64) umagicData { return umagic(64, c) } 138 139 // For signed division, we use a similar strategy. 140 // First, we enforce a positive c. 141 // x / c = -(x / (-c)) 142 // This will require an additional Neg op for c<0. 143 // 144 // If x is positive we're in a very similar state 145 // to the unsigned case above. We define: 146 // s = ⎡log2(c)⎤-1 147 // m = ⎡2^(n+s)/c⎤ 148 // Then 149 // ⎣x / c⎦ = ⎣x * m / 2^(n+s)⎦ 150 // If x is negative we have 151 // ⎡x / c⎤ = ⎣x * m / 2^(n+s)⎦ + 1 152 // (TODO: derivation?) 153 // 154 // The multiply is a bit odd, as it is a signed n-bit value 155 // times an unsigned n-bit value. For n smaller than the 156 // word size, we can extend x and m appropriately and use the 157 // signed multiply instruction. For n == word size, 158 // we must use the signed multiply high and correct 159 // the result by adding x*2^n. 160 // 161 // Adding 1 if x<0 is done by subtracting x>>(n-1). 162 163 func smagicOK(n uint, c int64) bool { 164 if c < 0 { 165 // Doesn't work for negative c. 166 return false 167 } 168 // Doesn't work for 0. 169 // Don't use it for powers of 2. 170 return c&(c-1) != 0 171 } 172 173 // smagicOKn reports whether we should strength reduce a signed n-bit divide by c. 174 func smagicOK8(c int8) bool { return smagicOK(8, int64(c)) } 175 func smagicOK16(c int16) bool { return smagicOK(16, int64(c)) } 176 func smagicOK32(c int32) bool { return smagicOK(32, int64(c)) } 177 func smagicOK64(c int64) bool { return smagicOK(64, c) } 178 179 type smagicData struct { 180 s int64 // ⎡log2(c)⎤-1 181 m uint64 // ⎡2^(n+s)/c⎤ 182 } 183 184 // smagic computes the constants needed to strength reduce signed n-bit divides by the constant c. 185 // Must have c>0. 186 // The return values satisfy for all -2^(n-1) <= x < 2^(n-1) 187 // 188 // trunc(x / c) = x * m >> (n+s) + (x < 0 ? 1 : 0) 189 func smagic(n uint, c int64) smagicData { 190 C := new(big.Int).SetInt64(c) 191 s := C.BitLen() - 1 192 M := big.NewInt(1) 193 M.Lsh(M, n+uint(s)) // 2^(n+s) 194 M.Add(M, C) // 2^(n+s)+c 195 M.Sub(M, big.NewInt(1)) // 2^(n+s)+c-1 196 M.Div(M, C) // ⎡2^(n+s)/c⎤ 197 if M.Bit(int(n)) != 0 { 198 panic("n+1st bit is set") 199 } 200 if M.Bit(int(n-1)) == 0 { 201 panic("nth bit is not set") 202 } 203 m := M.Uint64() 204 return smagicData{s: int64(s), m: m} 205 } 206 207 func smagic8(c int8) smagicData { return smagic(8, int64(c)) } 208 func smagic16(c int16) smagicData { return smagic(16, int64(c)) } 209 func smagic32(c int32) smagicData { return smagic(32, int64(c)) } 210 func smagic64(c int64) smagicData { return smagic(64, c) } 211 212 // Divisibility x%c == 0 can be checked more efficiently than directly computing 213 // the modulus x%c and comparing against 0. 214 // 215 // The same "Division by invariant integers using multiplication" paper 216 // by Granlund and Montgomery referenced above briefly mentions this method 217 // and it is further elaborated in "Hacker's Delight" by Warren Section 10-17 218 // 219 // The first thing to note is that for odd integers, exact division can be computed 220 // by using the modular inverse with respect to the word size 2^n. 221 // 222 // Given c, compute m such that (c * m) mod 2^n == 1 223 // Then if c divides x (x%c ==0), the quotient is given by q = x/c == x*m mod 2^n 224 // 225 // x can range from 0, c, 2c, 3c, ... ⎣(2^n - 1)/c⎦ * c the maximum multiple 226 // Thus, x*m mod 2^n is 0, 1, 2, 3, ... ⎣(2^n - 1)/c⎦ 227 // i.e. the quotient takes all values from zero up to max = ⎣(2^n - 1)/c⎦ 228 // 229 // If x is not divisible by c, then x*m mod 2^n must take some larger value than max. 230 // 231 // This gives x*m mod 2^n <= ⎣(2^n - 1)/c⎦ as a test for divisibility 232 // involving one multiplication and compare. 233 // 234 // To extend this to even integers, consider c = d0 * 2^k where d0 is odd. 235 // We can test whether x is divisible by both d0 and 2^k. 236 // For d0, the test is the same as above. Let m be such that m*d0 mod 2^n == 1 237 // Then x*m mod 2^n <= ⎣(2^n - 1)/d0⎦ is the first test. 238 // The test for divisibility by 2^k is a check for k trailing zeroes. 239 // Note that since d0 is odd, m is odd and thus x*m will have the same number of 240 // trailing zeroes as x. So the two tests are, 241 // 242 // x*m mod 2^n <= ⎣(2^n - 1)/d0⎦ 243 // and x*m ends in k zero bits 244 // 245 // These can be combined into a single comparison by the following 246 // (theorem ZRU in Hacker's Delight) for unsigned integers. 247 // 248 // x <= a and x ends in k zero bits if and only if RotRight(x ,k) <= ⎣a/(2^k)⎦ 249 // Where RotRight(x ,k) is right rotation of x by k bits. 250 // 251 // To prove the first direction, x <= a -> ⎣x/(2^k)⎦ <= ⎣a/(2^k)⎦ 252 // But since x ends in k zeroes all the rotated bits would be zero too. 253 // So RotRight(x, k) == ⎣x/(2^k)⎦ <= ⎣a/(2^k)⎦ 254 // 255 // If x does not end in k zero bits, then RotRight(x, k) 256 // has some non-zero bits in the k highest bits. 257 // ⎣x/(2^k)⎦ has all zeroes in the k highest bits, 258 // so RotRight(x, k) > ⎣x/(2^k)⎦ 259 // 260 // Finally, if x > a and has k trailing zero bits, then RotRight(x, k) == ⎣x/(2^k)⎦ 261 // and ⎣x/(2^k)⎦ must be greater than ⎣a/(2^k)⎦, that is the top n-k bits of x must 262 // be greater than the top n-k bits of a because the rest of x bits are zero. 263 // 264 // So the two conditions about can be replaced with the single test 265 // 266 // RotRight(x*m mod 2^n, k) <= ⎣(2^n - 1)/c⎦ 267 // 268 // Where d0*2^k was replaced by c on the right hand side. 269 270 // udivisibleOK reports whether we should strength reduce an unsigned n-bit divisibilty check by c. 271 func udivisibleOK(n uint, c int64) bool { 272 // Convert from ConstX auxint values to the real uint64 constant they represent. 273 d := uint64(c) << (64 - n) >> (64 - n) 274 275 // Doesn't work for 0. 276 // Don't use for powers of 2. 277 return d&(d-1) != 0 278 } 279 280 func udivisibleOK8(c int8) bool { return udivisibleOK(8, int64(c)) } 281 func udivisibleOK16(c int16) bool { return udivisibleOK(16, int64(c)) } 282 func udivisibleOK32(c int32) bool { return udivisibleOK(32, int64(c)) } 283 func udivisibleOK64(c int64) bool { return udivisibleOK(64, c) } 284 285 type udivisibleData struct { 286 k int64 // trailingZeros(c) 287 m uint64 // m * (c>>k) mod 2^n == 1 multiplicative inverse of odd portion modulo 2^n 288 max uint64 // ⎣(2^n - 1)/ c⎦ max value to for divisibility 289 } 290 291 func udivisible(n uint, c int64) udivisibleData { 292 // Convert from ConstX auxint values to the real uint64 constant they represent. 293 d := uint64(c) << (64 - n) >> (64 - n) 294 295 k := bits.TrailingZeros64(d) 296 d0 := d >> uint(k) // the odd portion of the divisor 297 298 mask := ^uint64(0) >> (64 - n) 299 300 // Calculate the multiplicative inverse via Newton's method. 301 // Quadratic convergence doubles the number of correct bits per iteration. 302 m := d0 // initial guess correct to 3-bits d0*d0 mod 8 == 1 303 m = m * (2 - m*d0) // 6-bits 304 m = m * (2 - m*d0) // 12-bits 305 m = m * (2 - m*d0) // 24-bits 306 m = m * (2 - m*d0) // 48-bits 307 m = m * (2 - m*d0) // 96-bits >= 64-bits 308 m = m & mask 309 310 max := mask / d 311 312 return udivisibleData{ 313 k: int64(k), 314 m: m, 315 max: max, 316 } 317 } 318 319 func udivisible8(c int8) udivisibleData { return udivisible(8, int64(c)) } 320 func udivisible16(c int16) udivisibleData { return udivisible(16, int64(c)) } 321 func udivisible32(c int32) udivisibleData { return udivisible(32, int64(c)) } 322 func udivisible64(c int64) udivisibleData { return udivisible(64, c) } 323 324 // For signed integers, a similar method follows. 325 // 326 // Given c > 1 and odd, compute m such that (c * m) mod 2^n == 1 327 // Then if c divides x (x%c ==0), the quotient is given by q = x/c == x*m mod 2^n 328 // 329 // x can range from ⎡-2^(n-1)/c⎤ * c, ... -c, 0, c, ... ⎣(2^(n-1) - 1)/c⎦ * c 330 // Thus, x*m mod 2^n is ⎡-2^(n-1)/c⎤, ... -2, -1, 0, 1, 2, ... ⎣(2^(n-1) - 1)/c⎦ 331 // 332 // So, x is a multiple of c if and only if: 333 // ⎡-2^(n-1)/c⎤ <= x*m mod 2^n <= ⎣(2^(n-1) - 1)/c⎦ 334 // 335 // Since c > 1 and odd, this can be simplified by 336 // ⎡-2^(n-1)/c⎤ == ⎡(-2^(n-1) + 1)/c⎤ == -⎣(2^(n-1) - 1)/c⎦ 337 // 338 // -⎣(2^(n-1) - 1)/c⎦ <= x*m mod 2^n <= ⎣(2^(n-1) - 1)/c⎦ 339 // 340 // To extend this to even integers, consider c = d0 * 2^k where d0 is odd. 341 // We can test whether x is divisible by both d0 and 2^k. 342 // 343 // Let m be such that (d0 * m) mod 2^n == 1. 344 // Let q = x*m mod 2^n. Then c divides x if: 345 // 346 // -⎣(2^(n-1) - 1)/d0⎦ <= q <= ⎣(2^(n-1) - 1)/d0⎦ and q ends in at least k 0-bits 347 // 348 // To transform this to a single comparison, we use the following theorem (ZRS in Hacker's Delight). 349 // 350 // For a >= 0 the following conditions are equivalent: 351 // 1) -a <= x <= a and x ends in at least k 0-bits 352 // 2) RotRight(x+a', k) <= ⎣2a'/2^k⎦ 353 // 354 // Where a' = a & -2^k (a with its right k bits set to zero) 355 // 356 // To see that 1 & 2 are equivalent, note that -a <= x <= a is equivalent to 357 // -a' <= x <= a' if and only if x ends in at least k 0-bits. Adding -a' to each side gives, 358 // 0 <= x + a' <= 2a' and x + a' ends in at least k 0-bits if and only if x does since a' has 359 // k 0-bits by definition. We can use theorem ZRU above with x -> x + a' and a -> 2a' giving 1) == 2). 360 // 361 // Let m be such that (d0 * m) mod 2^n == 1. 362 // Let q = x*m mod 2^n. 363 // Let a' = ⎣(2^(n-1) - 1)/d0⎦ & -2^k 364 // 365 // Then the divisibility test is: 366 // 367 // RotRight(q+a', k) <= ⎣2a'/2^k⎦ 368 // 369 // Note that the calculation is performed using unsigned integers. 370 // Since a' can have n-1 bits, 2a' may have n bits and there is no risk of overflow. 371 372 // sdivisibleOK reports whether we should strength reduce a signed n-bit divisibilty check by c. 373 func sdivisibleOK(n uint, c int64) bool { 374 if c < 0 { 375 // Doesn't work for negative c. 376 return false 377 } 378 // Doesn't work for 0. 379 // Don't use it for powers of 2. 380 return c&(c-1) != 0 381 } 382 383 func sdivisibleOK8(c int8) bool { return sdivisibleOK(8, int64(c)) } 384 func sdivisibleOK16(c int16) bool { return sdivisibleOK(16, int64(c)) } 385 func sdivisibleOK32(c int32) bool { return sdivisibleOK(32, int64(c)) } 386 func sdivisibleOK64(c int64) bool { return sdivisibleOK(64, c) } 387 388 type sdivisibleData struct { 389 k int64 // trailingZeros(c) 390 m uint64 // m * (c>>k) mod 2^n == 1 multiplicative inverse of odd portion modulo 2^n 391 a uint64 // ⎣(2^(n-1) - 1)/ (c>>k)⎦ & -(1<<k) additive constant 392 max uint64 // ⎣(2 a) / (1<<k)⎦ max value to for divisibility 393 } 394 395 func sdivisible(n uint, c int64) sdivisibleData { 396 d := uint64(c) 397 k := bits.TrailingZeros64(d) 398 d0 := d >> uint(k) // the odd portion of the divisor 399 400 mask := ^uint64(0) >> (64 - n) 401 402 // Calculate the multiplicative inverse via Newton's method. 403 // Quadratic convergence doubles the number of correct bits per iteration. 404 m := d0 // initial guess correct to 3-bits d0*d0 mod 8 == 1 405 m = m * (2 - m*d0) // 6-bits 406 m = m * (2 - m*d0) // 12-bits 407 m = m * (2 - m*d0) // 24-bits 408 m = m * (2 - m*d0) // 48-bits 409 m = m * (2 - m*d0) // 96-bits >= 64-bits 410 m = m & mask 411 412 a := ((mask >> 1) / d0) & -(1 << uint(k)) 413 max := (2 * a) >> uint(k) 414 415 return sdivisibleData{ 416 k: int64(k), 417 m: m, 418 a: a, 419 max: max, 420 } 421 } 422 423 func sdivisible8(c int8) sdivisibleData { return sdivisible(8, int64(c)) } 424 func sdivisible16(c int16) sdivisibleData { return sdivisible(16, int64(c)) } 425 func sdivisible32(c int32) sdivisibleData { return sdivisible(32, int64(c)) } 426 func sdivisible64(c int64) sdivisibleData { return sdivisible(64, c) } 427