// Copyright 2025 The Go Authors. All rights reserved. // Use of this source code is governed by a BSD-style // license that can be found in the LICENSE file. package strconv import "math/bits" var uint64pow10 = [...]uint64{ 1, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9, 1e10, 1e11, 1e12, 1e13, 1e14, 1e15, 1e16, 1e17, 1e18, 1e19, } // fixedFtoa formats a number of decimal digits of mant*(2^exp) into d, // where mant > 0 and 1 ≤ digits ≤ 18. // If fmt == 'f', digits is a conservative overestimate, and the final // number of digits is prec past the decimal point. func fixedFtoa(d *decimalSlice, mant uint64, exp, digits, prec int, fmt byte) { // The strategy here is to multiply (mant * 2^exp) by a power of 10 // to make the resulting integer be the number of digits we want. // // Adams proved in the Ryu paper that 128-bit precision in the // power-of-10 constant is sufficient to produce correctly // rounded output for all float64s, up to 18 digits. // https://dl.acm.org/doi/10.1145/3192366.3192369 // // TODO(rsc): The paper is not focused on, nor terribly clear about, // this fact in this context, and the proof seems too complicated. // Post a shorter, more direct proof and link to it here. if digits > 18 { panic("fixedFtoa called with digits > 18") } // Shift mantissa to have 64 bits, // so that the 192-bit product below will // have at least 63 bits in its top word. b := 64 - bits.Len64(mant) mant <<= b exp -= b // We have f = mant * 2^exp ≥ 2^(63+exp) // and we want to multiply it by some 10^p // to make it have the number of digits plus one rounding bit: // // 2 * 10^(digits-1) ≤ f * 10^p < ~2 * 10^digits // // The lower bound is required, but the upper bound is approximate: // we must not have too few digits, but we can round away extra ones. // // f * 10^p ≥ 2 * 10^(digits-1) // 10^p ≥ 2 * 10^(digits-1) / f [dividing by f] // p ≥ (log₁₀ 2) + (digits-1) - log₁₀ f [taking log₁₀] // p ≥ (log₁₀ 2) + (digits-1) - log₁₀ (mant * 2^exp) [expanding f] // p ≥ (log₁₀ 2) + (digits-1) - (log₁₀ 2) * (64 + exp) [mant < 2⁶⁴] // p ≥ (digits - 1) - (log₁₀ 2) * (63 + exp) [refactoring] // // Once we have p, we can compute the scaled value: // // dm * 2^de = mant * 2^exp * 10^p // = mant * 2^exp * pow/2^128 * 2^exp2. // = (mant * pow/2^128) * 2^(exp+exp2). p := (digits - 1) - mulLog10_2(63+exp) pow, exp2, ok := pow10(p) if !ok { // This never happens due to the range of float32/float64 exponent panic("fixedFtoa: pow10 out of range") } if -22 <= p && p < 0 { // Special case: Let q=-p. q is in [1,22]. We are dividing by 10^q // and the mantissa may be a multiple of 5^q (5^22 < 2^53), // in which case the division must be computed exactly and // recorded as exact for correct rounding. Our normal computation is: // // dm = floor(mant * floor(10^p * 2^s)) // // for some scaling shift s. To make this an exact division, // it suffices to change the inner floor to a ceil: // // dm = floor(mant * ceil(10^p * 2^s)) // // In the range of values we are using, the floor and ceil // cancel each other out and the high 64 bits of the product // come out exactly right. // (This is the same trick compilers use for division by constants. // See Hacker's Delight, 2nd ed., Chapter 10.) pow.Lo++ } dm, lo1, lo0 := umul192(mant, pow) de := exp + exp2 // Check whether any bits have been truncated from dm. // If so, set dt != 0. If not, leave dt == 0 (meaning dm is exact). var dt uint switch { default: // Most powers of 10 use a truncated constant, // meaning the result is also truncated. dt = 1 case 0 <= p && p <= 55: // Small positive powers of 10 (up to 10⁵⁵) can be represented // precisely in a 128-bit mantissa (5⁵⁵ ≤ 2¹²⁸), so the only truncation // comes from discarding the low bits of the 192-bit product. // // TODO(rsc): The new proof mentioned above should also // prove that we can't have lo1 == 0 and lo0 != 0. // After proving that, drop computation and use of lo0 here. dt = bool2uint(lo1|lo0 != 0) case -22 <= p && p < 0 && divisiblePow5(mant, -p): // If the original mantissa was a multiple of 5^p, // the result is exact. (See comment above for pow.Lo++.) dt = 0 } // The value we want to format is dm * 2^de, where de < 0. // Multply by 2^de by shifting, but leave one extra bit for rounding. // After the shift, the "integer part" of dm is dm>>1, // the "rounding bit" (the first fractional bit) is dm&1, // and the "truncated bit" (have any bits been discarded?) is dt. shift := -de - 1 dt |= bool2uint(dm&(1<>= shift // Set decimal point in eventual formatted digits, // so we can update it as we adjust the digits. d.dp = digits - p // Trim excess digit if any, updating truncation and decimal point. // The << 1 is leaving room for the rounding bit. max := uint64pow10[digits] << 1 if dm >= max { var r uint dm, r = dm/10, uint(dm%10) dt |= bool2uint(r != 0) d.dp++ } // If this is %.*f we may have overestimated the digits needed. // Now that we know where the decimal point is, // trim to the actual number of digits, which is d.dp+prec. if fmt == 'f' && digits != d.dp+prec { for digits > d.dp+prec { var r uint dm, r = dm/10, uint(dm%10) dt |= bool2uint(r != 0) digits-- } // Dropping those digits can create a new leftmost // non-zero digit, like if we are formatting %.1f and // convert 0.09 -> 0.1. Detect and adjust for that. if digits <= 0 { digits = 1 d.dp++ } max = uint64pow10[digits] << 1 } // Round and shift away rounding bit. // We want to round up when // (a) the fractional part is > 0.5 (dm&1 != 0 and dt == 1) // (b) or the fractional part is ≥ 0.5 and the integer part is odd // (dm&1 != 0 and dm&2 != 0). // The bitwise expression encodes that logic. dm += uint64(uint(dm) & (dt | uint(dm)>>1) & 1) dm >>= 1 if dm == max>>1 { // 999... rolled over to 1000... dm = uint64pow10[digits-1] d.dp++ } // Format digits into d. if dm != 0 { if formatBase10(d.d[:digits], dm) != 0 { panic("formatBase10") } d.nd = digits for d.d[d.nd-1] == '0' { d.nd-- } } }