Source file src/cmd/compile/internal/ssa/magic.go

Documentation: cmd/compile/internal/ssa

     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  type umagicData struct {
   100  	s int64  // ⎡log2(c)⎤
   101  	m uint64 // ⎡2^(n+s)/c⎤ - 2^n
   102  }
   103  
   104  // umagic computes the constants needed to strength reduce unsigned n-bit divides by the constant uint64(c).
   105  // The return values satisfy for all 0 <= x < 2^n
   106  //  floor(x / uint64(c)) = x * (m + 2^n) >> (n+s)
   107  func umagic(n uint, c int64) umagicData {
   108  	// Convert from ConstX auxint values to the real uint64 constant they represent.
   109  	d := uint64(c) << (64 - n) >> (64 - n)
   110  
   111  	C := new(big.Int).SetUint64(d)
   112  	s := C.BitLen()
   113  	M := big.NewInt(1)
   114  	M.Lsh(M, n+uint(s))     // 2^(n+s)
   115  	M.Add(M, C)             // 2^(n+s)+c
   116  	M.Sub(M, big.NewInt(1)) // 2^(n+s)+c-1
   117  	M.Div(M, C)             // ⎡2^(n+s)/c⎤
   118  	if M.Bit(int(n)) != 1 {
   119  		panic("n+1st bit isn't set")
   120  	}
   121  	M.SetBit(M, int(n), 0)
   122  	m := M.Uint64()
   123  	return umagicData{s: int64(s), m: m}
   124  }
   125  
   126  // For signed division, we use a similar strategy.
   127  // First, we enforce a positive c.
   128  //   x / c = -(x / (-c))
   129  // This will require an additional Neg op for c<0.
   130  //
   131  // If x is positive we're in a very similar state
   132  // to the unsigned case above.  We define:
   133  //   s = ⎡log2(c)⎤-1
   134  //   m = ⎡2^(n+s)/c⎤
   135  // Then
   136  //   ⎣x / c⎦ = ⎣x * m / 2^(n+s)⎦
   137  // If x is negative we have
   138  //   ⎡x / c⎤ = ⎣x * m / 2^(n+s)⎦ + 1
   139  // (TODO: derivation?)
   140  //
   141  // The multiply is a bit odd, as it is a signed n-bit value
   142  // times an unsigned n-bit value.  For n smaller than the
   143  // word size, we can extend x and m appropriately and use the
   144  // signed multiply instruction.  For n == word size,
   145  // we must use the signed multiply high and correct
   146  // the result by adding x*2^n.
   147  //
   148  // Adding 1 if x<0 is done by subtracting x>>(n-1).
   149  
   150  func smagicOK(n uint, c int64) bool {
   151  	if c < 0 {
   152  		// Doesn't work for negative c.
   153  		return false
   154  	}
   155  	// Doesn't work for 0.
   156  	// Don't use it for powers of 2.
   157  	return c&(c-1) != 0
   158  }
   159  
   160  type smagicData struct {
   161  	s int64  // ⎡log2(c)⎤-1
   162  	m uint64 // ⎡2^(n+s)/c⎤
   163  }
   164  
   165  // magic computes the constants needed to strength reduce signed n-bit divides by the constant c.
   166  // Must have c>0.
   167  // The return values satisfy for all -2^(n-1) <= x < 2^(n-1)
   168  //  trunc(x / c) = x * m >> (n+s) + (x < 0 ? 1 : 0)
   169  func smagic(n uint, c int64) smagicData {
   170  	C := new(big.Int).SetInt64(c)
   171  	s := C.BitLen() - 1
   172  	M := big.NewInt(1)
   173  	M.Lsh(M, n+uint(s))     // 2^(n+s)
   174  	M.Add(M, C)             // 2^(n+s)+c
   175  	M.Sub(M, big.NewInt(1)) // 2^(n+s)+c-1
   176  	M.Div(M, C)             // ⎡2^(n+s)/c⎤
   177  	if M.Bit(int(n)) != 0 {
   178  		panic("n+1st bit is set")
   179  	}
   180  	if M.Bit(int(n-1)) == 0 {
   181  		panic("nth bit is not set")
   182  	}
   183  	m := M.Uint64()
   184  	return smagicData{s: int64(s), m: m}
   185  }
   186  
   187  // Divisibility x%c == 0 can be checked more efficiently than directly computing
   188  // the modulus x%c and comparing against 0.
   189  //
   190  // The same "Division by invariant integers using multiplication" paper
   191  // by Granlund and Montgomery referenced above briefly mentions this method
   192  // and it is further elaborated in "Hacker's Delight" by Warren Section 10-17
   193  //
   194  // The first thing to note is that for odd integers, exact division can be computed
   195  // by using the modular inverse with respect to the word size 2^n.
   196  //
   197  // Given c, compute m such that (c * m) mod 2^n == 1
   198  // Then if c divides x (x%c ==0), the quotient is given by q = x/c == x*m mod 2^n
   199  //
   200  // x can range from 0, c, 2c, 3c, ... ⎣(2^n - 1)/c⎦ * c the maximum multiple
   201  // Thus, x*m mod 2^n is 0, 1, 2, 3, ... ⎣(2^n - 1)/c⎦
   202  // i.e. the quotient takes all values from zero up to max = ⎣(2^n - 1)/c⎦
   203  //
   204  // If x is not divisible by c, then x*m mod 2^n must take some larger value than max.
   205  //
   206  // This gives x*m mod 2^n <= ⎣(2^n - 1)/c⎦ as a test for divisibility
   207  // involving one multiplication and compare.
   208  //
   209  // To extend this to even integers, consider c = d0 * 2^k where d0 is odd.
   210  // We can test whether x is divisible by both d0 and 2^k.
   211  // For d0, the test is the same as above.  Let m be such that m*d0 mod 2^n == 1
   212  // Then x*m mod 2^n <= ⎣(2^n - 1)/d0⎦ is the first test.
   213  // The test for divisibility by 2^k is a check for k trailing zeroes.
   214  // Note that since d0 is odd, m is odd and thus x*m will have the same number of
   215  // trailing zeroes as x.  So the two tests are,
   216  //
   217  // x*m mod 2^n <= ⎣(2^n - 1)/d0⎦
   218  // and x*m ends in k zero bits
   219  //
   220  // These can be combined into a single comparison by the following
   221  // (theorem ZRU in Hacker's Delight) for unsigned integers.
   222  //
   223  // x <= a and x ends in k zero bits if and only if RotRight(x ,k) <= ⎣a/(2^k)⎦
   224  // Where RotRight(x ,k) is right rotation of x by k bits.
   225  //
   226  // To prove the first direction, x <= a -> ⎣x/(2^k)⎦ <= ⎣a/(2^k)⎦
   227  // But since x ends in k zeroes all the rotated bits would be zero too.
   228  // So RotRight(x, k) == ⎣x/(2^k)⎦ <= ⎣a/(2^k)⎦
   229  //
   230  // If x does not end in k zero bits, then RotRight(x, k)
   231  // has some non-zero bits in the k highest bits.
   232  // ⎣x/(2^k)⎦ has all zeroes in the k highest bits,
   233  // so RotRight(x, k) > ⎣x/(2^k)⎦
   234  //
   235  // Finally, if x > a and has k trailing zero bits, then RotRight(x, k) == ⎣x/(2^k)⎦
   236  // and ⎣x/(2^k)⎦ must be greater than ⎣a/(2^k)⎦, that is the top n-k bits of x must
   237  // be greater than the top n-k bits of a because the rest of x bits are zero.
   238  //
   239  // So the two conditions about can be replaced with the single test
   240  //
   241  // RotRight(x*m mod 2^n, k) <= ⎣(2^n - 1)/c⎦
   242  //
   243  // Where d0*2^k was replaced by c on the right hand side.
   244  
   245  // uivisibleOK reports whether we should strength reduce a n-bit dividisibilty check by c.
   246  func udivisibleOK(n uint, c int64) bool {
   247  	// Convert from ConstX auxint values to the real uint64 constant they represent.
   248  	d := uint64(c) << (64 - n) >> (64 - n)
   249  
   250  	// Doesn't work for 0.
   251  	// Don't use for powers of 2.
   252  	return d&(d-1) != 0
   253  }
   254  
   255  type udivisibleData struct {
   256  	k   int64  // trailingZeros(c)
   257  	m   uint64 // m * (c>>k) mod 2^n == 1 multiplicative inverse of odd portion modulo 2^n
   258  	max uint64 // ⎣(2^n - 1)/ c⎦ max value to for divisibility
   259  }
   260  
   261  func udivisible(n uint, c int64) udivisibleData {
   262  	// Convert from ConstX auxint values to the real uint64 constant they represent.
   263  	d := uint64(c) << (64 - n) >> (64 - n)
   264  
   265  	k := bits.TrailingZeros64(d)
   266  	d0 := d >> uint(k) // the odd portion of the divisor
   267  
   268  	mask := ^uint64(0) >> (64 - n)
   269  
   270  	// Calculate the multiplicative inverse via Newton's method.
   271  	// Quadratic convergence doubles the number of correct bits per iteration.
   272  	m := d0            // initial guess correct to 3-bits d0*d0 mod 8 == 1
   273  	m = m * (2 - m*d0) // 6-bits
   274  	m = m * (2 - m*d0) // 12-bits
   275  	m = m * (2 - m*d0) // 24-bits
   276  	m = m * (2 - m*d0) // 48-bits
   277  	m = m * (2 - m*d0) // 96-bits >= 64-bits
   278  	m = m & mask
   279  
   280  	max := mask / d
   281  
   282  	return udivisibleData{
   283  		k:   int64(k),
   284  		m:   m,
   285  		max: max,
   286  	}
   287  }
   288  
   289  // For signed integers, a similar method follows.
   290  //
   291  // Given c > 1 and odd, compute m such that (c * m) mod 2^n == 1
   292  // Then if c divides x (x%c ==0), the quotient is given by q = x/c == x*m mod 2^n
   293  //
   294  // x can range from ⎡-2^(n-1)/c⎤ * c, ... -c, 0, c, ...  ⎣(2^(n-1) - 1)/c⎦ * c
   295  // Thus, x*m mod 2^n is ⎡-2^(n-1)/c⎤, ... -2, -1, 0, 1, 2, ... ⎣(2^(n-1) - 1)/c⎦
   296  //
   297  // So, x is a multiple of c if and only if:
   298  // ⎡-2^(n-1)/c⎤ <= x*m mod 2^n <= ⎣(2^(n-1) - 1)/c⎦
   299  //
   300  // Since c > 1 and odd, this can be simplified by
   301  // ⎡-2^(n-1)/c⎤ == ⎡(-2^(n-1) + 1)/c⎤ == -⎣(2^(n-1) - 1)/c⎦
   302  //
   303  // -⎣(2^(n-1) - 1)/c⎦ <= x*m mod 2^n <= ⎣(2^(n-1) - 1)/c⎦
   304  //
   305  // To extend this to even integers, consider c = d0 * 2^k where d0 is odd.
   306  // We can test whether x is divisible by both d0 and 2^k.
   307  //
   308  // Let m be such that (d0 * m) mod 2^n == 1.
   309  // Let q = x*m mod 2^n. Then c divides x if:
   310  //
   311  // -⎣(2^(n-1) - 1)/d0⎦ <= q <= ⎣(2^(n-1) - 1)/d0⎦ and q ends in at least k 0-bits
   312  //
   313  // To transform this to a single comparison, we use the following theorem (ZRS in Hacker's Delight).
   314  //
   315  // For a >= 0 the following conditions are equivalent:
   316  // 1) -a <= x <= a and x ends in at least k 0-bits
   317  // 2) RotRight(x+a', k) <= ⎣2a'/2^k⎦
   318  //
   319  // Where a' = a & -2^k (a with its right k bits set to zero)
   320  //
   321  // To see that 1 & 2 are equivalent, note that -a <= x <= a is equivalent to
   322  // -a' <= x <= a' if and only if x ends in at least k 0-bits.  Adding -a' to each side gives,
   323  // 0 <= x + a' <= 2a' and x + a' ends in at least k 0-bits if and only if x does since a' has
   324  // k 0-bits by definition.  We can use theorem ZRU above with x -> x + a' and a -> 2a' giving 1) == 2).
   325  //
   326  // Let m be such that (d0 * m) mod 2^n == 1.
   327  // Let q = x*m mod 2^n.
   328  // Let a' = ⎣(2^(n-1) - 1)/d0⎦ & -2^k
   329  //
   330  // Then the divisibility test is:
   331  //
   332  // RotRight(q+a', k) <= ⎣2a'/2^k⎦
   333  //
   334  // Note that the calculation is performed using unsigned integers.
   335  // Since a' can have n-1 bits, 2a' may have n bits and there is no risk of overflow.
   336  
   337  // sdivisibleOK reports whether we should strength reduce a n-bit dividisibilty check by c.
   338  func sdivisibleOK(n uint, c int64) bool {
   339  	if c < 0 {
   340  		// Doesn't work for negative c.
   341  		return false
   342  	}
   343  	// Doesn't work for 0.
   344  	// Don't use it for powers of 2.
   345  	return c&(c-1) != 0
   346  }
   347  
   348  type sdivisibleData struct {
   349  	k   int64  // trailingZeros(c)
   350  	m   uint64 // m * (c>>k) mod 2^n == 1 multiplicative inverse of odd portion modulo 2^n
   351  	a   uint64 // ⎣(2^(n-1) - 1)/ (c>>k)⎦ & -(1<<k) additive constant
   352  	max uint64 // ⎣(2 a) / (1<<k)⎦ max value to for divisibility
   353  }
   354  
   355  func sdivisible(n uint, c int64) sdivisibleData {
   356  	d := uint64(c)
   357  	k := bits.TrailingZeros64(d)
   358  	d0 := d >> uint(k) // the odd portion of the divisor
   359  
   360  	mask := ^uint64(0) >> (64 - n)
   361  
   362  	// Calculate the multiplicative inverse via Newton's method.
   363  	// Quadratic convergence doubles the number of correct bits per iteration.
   364  	m := d0            // initial guess correct to 3-bits d0*d0 mod 8 == 1
   365  	m = m * (2 - m*d0) // 6-bits
   366  	m = m * (2 - m*d0) // 12-bits
   367  	m = m * (2 - m*d0) // 24-bits
   368  	m = m * (2 - m*d0) // 48-bits
   369  	m = m * (2 - m*d0) // 96-bits >= 64-bits
   370  	m = m & mask
   371  
   372  	a := ((mask >> 1) / d0) & -(1 << uint(k))
   373  	max := (2 * a) >> uint(k)
   374  
   375  	return sdivisibleData{
   376  		k:   int64(k),
   377  		m:   m,
   378  		a:   a,
   379  		max: max,
   380  	}
   381  }
   382  

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