gh-ost/vendor/github.com/shopspring/decimal/rounding.go
Tim Vaillancourt 47d49c6b92
Add go mod (#935)
* Add a go.mod file

* run go mod vendor again

* Move to a well-supported ini file reader

* Remove GO111MODULE=off

* Use go 1.16

* Rename github.com/outbrain/golib -> github.com/openark/golib

* Remove *.go-e files

* Fix for `strconv.ParseInt: parsing "": invalid syntax` error

* Add test for '[osc]' section

Co-authored-by: Nate Wernimont <nate.wernimont@workiva.com>
2021-06-24 20:19:37 +02:00

119 lines
3.8 KiB
Go

// Copyright 2009 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.
// Multiprecision decimal numbers.
// For floating-point formatting only; not general purpose.
// Only operations are assign and (binary) left/right shift.
// Can do binary floating point in multiprecision decimal precisely
// because 2 divides 10; cannot do decimal floating point
// in multiprecision binary precisely.
package decimal
type floatInfo struct {
mantbits uint
expbits uint
bias int
}
var float32info = floatInfo{23, 8, -127}
var float64info = floatInfo{52, 11, -1023}
// roundShortest rounds d (= mant * 2^exp) to the shortest number of digits
// that will let the original floating point value be precisely reconstructed.
func roundShortest(d *decimal, mant uint64, exp int, flt *floatInfo) {
// If mantissa is zero, the number is zero; stop now.
if mant == 0 {
d.nd = 0
return
}
// Compute upper and lower such that any decimal number
// between upper and lower (possibly inclusive)
// will round to the original floating point number.
// We may see at once that the number is already shortest.
//
// Suppose d is not denormal, so that 2^exp <= d < 10^dp.
// The closest shorter number is at least 10^(dp-nd) away.
// The lower/upper bounds computed below are at distance
// at most 2^(exp-mantbits).
//
// So the number is already shortest if 10^(dp-nd) > 2^(exp-mantbits),
// or equivalently log2(10)*(dp-nd) > exp-mantbits.
// It is true if 332/100*(dp-nd) >= exp-mantbits (log2(10) > 3.32).
minexp := flt.bias + 1 // minimum possible exponent
if exp > minexp && 332*(d.dp-d.nd) >= 100*(exp-int(flt.mantbits)) {
// The number is already shortest.
return
}
// d = mant << (exp - mantbits)
// Next highest floating point number is mant+1 << exp-mantbits.
// Our upper bound is halfway between, mant*2+1 << exp-mantbits-1.
upper := new(decimal)
upper.Assign(mant*2 + 1)
upper.Shift(exp - int(flt.mantbits) - 1)
// d = mant << (exp - mantbits)
// Next lowest floating point number is mant-1 << exp-mantbits,
// unless mant-1 drops the significant bit and exp is not the minimum exp,
// in which case the next lowest is mant*2-1 << exp-mantbits-1.
// Either way, call it mantlo << explo-mantbits.
// Our lower bound is halfway between, mantlo*2+1 << explo-mantbits-1.
var mantlo uint64
var explo int
if mant > 1<<flt.mantbits || exp == minexp {
mantlo = mant - 1
explo = exp
} else {
mantlo = mant*2 - 1
explo = exp - 1
}
lower := new(decimal)
lower.Assign(mantlo*2 + 1)
lower.Shift(explo - int(flt.mantbits) - 1)
// The upper and lower bounds are possible outputs only if
// the original mantissa is even, so that IEEE round-to-even
// would round to the original mantissa and not the neighbors.
inclusive := mant%2 == 0
// Now we can figure out the minimum number of digits required.
// Walk along until d has distinguished itself from upper and lower.
for i := 0; i < d.nd; i++ {
l := byte('0') // lower digit
if i < lower.nd {
l = lower.d[i]
}
m := d.d[i] // middle digit
u := byte('0') // upper digit
if i < upper.nd {
u = upper.d[i]
}
// Okay to round down (truncate) if lower has a different digit
// or if lower is inclusive and is exactly the result of rounding
// down (i.e., and we have reached the final digit of lower).
okdown := l != m || inclusive && i+1 == lower.nd
// Okay to round up if upper has a different digit and either upper
// is inclusive or upper is bigger than the result of rounding up.
okup := m != u && (inclusive || m+1 < u || i+1 < upper.nd)
// If it's okay to do either, then round to the nearest one.
// If it's okay to do only one, do it.
switch {
case okdown && okup:
d.Round(i + 1)
return
case okdown:
d.RoundDown(i + 1)
return
case okup:
d.RoundUp(i + 1)
return
}
}
}