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| 1 | +// Copyright 2025 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 strconv |
| 6 | + |
| 7 | +import "math/bits" |
| 8 | + |
| 9 | +var uint64pow10 = [...]uint64{ |
| 10 | +1, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9, |
| 11 | +1e10, 1e11, 1e12, 1e13, 1e14, 1e15, 1e16, 1e17, 1e18, 1e19, |
| 12 | +} |
| 13 | + |
| 14 | +// fixedFtoa formats a number of decimal digits of mant*(2^exp) into d, |
| 15 | +// where mant > 0 and 1 ≤ digits ≤ 18. |
| 16 | +func fixedFtoa(d *decimalSlice, mant uint64, exp, digits int) { |
| 17 | +// The strategy here is to multiply (mant * 2^exp) by a power of 10 |
| 18 | +// to make the resulting integer be the number of digits we want. |
| 19 | +// |
| 20 | +// Adams proved in the Ryu paper that 128-bit precision in the |
| 21 | +// power-of-10 constant is sufficient to produce correctly |
| 22 | +// rounded output for all float64s, up to 18 digits. |
| 23 | +// https://dl.acm.org/doi/10.1145/3192366.3192369 |
| 24 | +// |
| 25 | +// TODO(rsc): The paper is not focused on, nor terribly clear about, |
| 26 | +// this fact in this context, and the proof seems too complicated. |
| 27 | +// Post a shorter, more direct proof and link to it here. |
| 28 | + |
| 29 | +if digits > 18 { |
| 30 | +panic("fixedFtoa called with digits > 18") |
| 31 | +} |
| 32 | + |
| 33 | +// Shift mantissa to have 64 bits, |
| 34 | +// so that the 192-bit product below will |
| 35 | +// have at least 63 bits in its top word. |
| 36 | +b := 64 - bits.Len64(mant) |
| 37 | +mant <<= b |
| 38 | +exp -= b |
| 39 | + |
| 40 | +// We have f = mant * 2^exp ≥ 2^(63+exp) |
| 41 | +// and we want to multiply it by some 10^p |
| 42 | +// to make it have the number of digits plus one rounding bit: |
| 43 | +// |
| 44 | +// 2 * 10^(digits-1) ≤ f * 10^p < ~2 * 10^digits |
| 45 | +// |
| 46 | +// The lower bound is required, but the upper bound is approximate: |
| 47 | +// we must not have too few digits, but we can round away extra ones. |
| 48 | +// |
| 49 | +// f * 10^p ≥ 2 * 10^(digits-1) |
| 50 | +// 10^p ≥ 2 * 10^(digits-1) / f [dividing by f] |
| 51 | +// p ≥ (log₁₀ 2) + (digits-1) - log₁₀ f [taking log₁₀] |
| 52 | +// p ≥ (log₁₀ 2) + (digits-1) - log₁₀ (mant * 2^exp) [expanding f] |
| 53 | +// p ≥ (log₁₀ 2) + (digits-1) - (log₁₀ 2) * (64 + exp) [mant < 2⁶⁴] |
| 54 | +// p ≥ (digits - 1) - (log₁₀ 2) * (63 + exp) [refactoring] |
| 55 | +// |
| 56 | +// Once we have p, we can compute the scaled value: |
| 57 | +// |
| 58 | +// dm * 2^de = mant * 2^exp * 10^p |
| 59 | +// = mant * 2^exp * pow/2^128 * 2^exp2. |
| 60 | +// = (mant * pow/2^128) * 2^(exp+exp2). |
| 61 | +p := (digits - 1) - mulLog10_2(63+exp) |
| 62 | +pow, exp2, ok := pow10(p) |
| 63 | +if !ok { |
| 64 | +// This never happens due to the range of float32/float64 exponent |
| 65 | +panic("fixedFtoa: pow10 out of range") |
| 66 | +} |
| 67 | +if -22 <= p && p < 0 { |
| 68 | +// Special case: Let q=-p. q is in [1,22]. We are dividing by 10^q |
| 69 | +// and the mantissa may be a multiple of 5^q (5^22 < 2^53), |
| 70 | +// in which case the division must be computed exactly and |
| 71 | +// recorded as exact for correct rounding. Our normal computation is: |
| 72 | +// |
| 73 | +// dm = floor(mant * floor(10^p * 2^s)) |
| 74 | +// |
| 75 | +// for some scaling shift s. To make this an exact division, |
| 76 | +// it suffices to change the inner floor to a ceil: |
| 77 | +// |
| 78 | +// dm = floor(mant * ceil(10^p * 2^s)) |
| 79 | +// |
| 80 | +// In the range of values we are using, the floor and ceil |
| 81 | +// cancel each other out and the high 64 bits of the product |
| 82 | +// come out exactly right. |
| 83 | +// (This is the same trick compilers use for division by constants. |
| 84 | +// See Hacker's Delight, 2nd ed., Chapter 10.) |
| 85 | +pow.Lo++ |
| 86 | +} |
| 87 | +dm, lo1, lo0 := umul192(mant, pow) |
| 88 | +de := exp + exp2 |
| 89 | + |
| 90 | +// Check whether any bits have been truncated from dm. |
| 91 | +// If so, set dt != 0. If not, leave dt == 0 (meaning dm is exact). |
| 92 | +var dt uint |
| 93 | +switch { |
| 94 | +default: |
| 95 | +// Most powers of 10 use a truncated constant, |
| 96 | +// meaning the result is also truncated. |
| 97 | +dt = 1 |
| 98 | +case 0 <= p && p <= 55: |
| 99 | +// Small positive powers of 10 (up to 10⁵⁵) can be represented |
| 100 | +// precisely in a 128-bit mantissa (5⁵⁵ ≤ 2¹²⁸), so the only truncation |
| 101 | +// comes from discarding the low bits of the 192-bit product. |
| 102 | +// |
| 103 | +// TODO(rsc): The new proof mentioned above should also |
| 104 | +// prove that we can't have lo1 == 0 and lo0 != 0. |
| 105 | +// After proving that, drop computation and use of lo0 here. |
| 106 | +dt = bool2uint(lo1|lo0 != 0) |
| 107 | +case -22 <= p && p < 0 && divisiblePow5(mant, -p): |
| 108 | +// If the original mantissa was a multiple of 5^p, |
| 109 | +// the result is exact. (See comment above for pow.Lo++.) |
| 110 | +dt = 0 |
| 111 | +} |
| 112 | + |
| 113 | +// The value we want to format is dm * 2^de, where de < 0. |
| 114 | +// Multply by 2^de by shifting, but leave one extra bit for rounding. |
| 115 | +// After the shift, the "integer part" of dm is dm>>1, |
| 116 | +// the "rounding bit" (the first fractional bit) is dm&1, |
| 117 | +// and the "truncated bit" (have any bits been discarded?) is dt. |
| 118 | +shift := -de - 1 |
| 119 | +dt |= bool2uint(dm&(1<<shift-1) != 0) |
| 120 | +dm >>= shift |
| 121 | + |
| 122 | +// Set decimal point in eventual formatted digits, |
| 123 | +// so we can update it as we adjust the digits. |
| 124 | +d.dp = digits - p |
| 125 | + |
| 126 | +// Trim excess digit if any, updating truncation and decimal point. |
| 127 | +// The << 1 is leaving room for the rounding bit. |
| 128 | +max := uint64pow10[digits] << 1 |
| 129 | +if dm >= max { |
| 130 | +var r uint |
| 131 | +dm, r = dm/10, uint(dm%10) |
| 132 | +dt |= bool2uint(r != 0) |
| 133 | +d.dp++ |
| 134 | +} |
| 135 | + |
| 136 | +// Round and shift away rounding bit. |
| 137 | +// We want to round up when |
| 138 | +// (a) the fractional part is > 0.5 (dm&1 != 0 and dt == 1) |
| 139 | +// (b) or the fractional part is ≥ 0.5 and the integer part is odd |
| 140 | +// (dm&1 != 0 and dm&2 != 0). |
| 141 | +// The bitwise expression encodes that logic. |
| 142 | +dm += uint64(uint(dm) & (dt | uint(dm)>>1) & 1) |
| 143 | +dm >>= 1 |
| 144 | +if dm == max>>1 { |
| 145 | +// 999... rolled over to 1000... |
| 146 | +dm = uint64pow10[digits-1] |
| 147 | +d.dp++ |
| 148 | +} |
| 149 | + |
| 150 | +// Format digits into d. |
| 151 | +formatBase10(d.d[:digits], dm) |
| 152 | +d.nd = digits |
| 153 | +for d.d[d.nd-1] == '0' { |
| 154 | +d.nd-- |
| 155 | +} |
| 156 | +} |
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