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			160 lines
		
	
	
	
		
			5.1 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
			
		
		
	
	
			160 lines
		
	
	
	
		
			5.1 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
| // Copyright 2009 The Go Authors. All rights reserved.
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| // Use of this source code is governed by a BSD-style
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| // license that can be found in the LICENSE file.
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| 
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| // Multiprecision decimal numbers.
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| // For floating-point formatting only; not general purpose.
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| // Only operations are assign and (binary) left/right shift.
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| // Can do binary floating point in multiprecision decimal precisely
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| // because 2 divides 10; cannot do decimal floating point
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| // in multiprecision binary precisely.
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| 
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| package decimal
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| 
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| type floatInfo struct {
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| 	mantbits uint
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| 	expbits  uint
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| 	bias     int
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| }
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| 
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| var float32info = floatInfo{23, 8, -127}
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| var float64info = floatInfo{52, 11, -1023}
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| 
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| // roundShortest rounds d (= mant * 2^exp) to the shortest number of digits
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| // that will let the original floating point value be precisely reconstructed.
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| func roundShortest(d *decimal, mant uint64, exp int, flt *floatInfo) {
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| 	// If mantissa is zero, the number is zero; stop now.
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| 	if mant == 0 {
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| 		d.nd = 0
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| 		return
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| 	}
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| 
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| 	// Compute upper and lower such that any decimal number
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| 	// between upper and lower (possibly inclusive)
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| 	// will round to the original floating point number.
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| 
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| 	// We may see at once that the number is already shortest.
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| 	//
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| 	// Suppose d is not denormal, so that 2^exp <= d < 10^dp.
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| 	// The closest shorter number is at least 10^(dp-nd) away.
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| 	// The lower/upper bounds computed below are at distance
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| 	// at most 2^(exp-mantbits).
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| 	//
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| 	// So the number is already shortest if 10^(dp-nd) > 2^(exp-mantbits),
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| 	// or equivalently log2(10)*(dp-nd) > exp-mantbits.
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| 	// It is true if 332/100*(dp-nd) >= exp-mantbits (log2(10) > 3.32).
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| 	minexp := flt.bias + 1 // minimum possible exponent
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| 	if exp > minexp && 332*(d.dp-d.nd) >= 100*(exp-int(flt.mantbits)) {
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| 		// The number is already shortest.
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| 		return
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| 	}
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| 
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| 	// d = mant << (exp - mantbits)
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| 	// Next highest floating point number is mant+1 << exp-mantbits.
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| 	// Our upper bound is halfway between, mant*2+1 << exp-mantbits-1.
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| 	upper := new(decimal)
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| 	upper.Assign(mant*2 + 1)
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| 	upper.Shift(exp - int(flt.mantbits) - 1)
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| 
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| 	// d = mant << (exp - mantbits)
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| 	// Next lowest floating point number is mant-1 << exp-mantbits,
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| 	// unless mant-1 drops the significant bit and exp is not the minimum exp,
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| 	// in which case the next lowest is mant*2-1 << exp-mantbits-1.
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| 	// Either way, call it mantlo << explo-mantbits.
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| 	// Our lower bound is halfway between, mantlo*2+1 << explo-mantbits-1.
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| 	var mantlo uint64
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| 	var explo int
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| 	if mant > 1<<flt.mantbits || exp == minexp {
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| 		mantlo = mant - 1
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| 		explo = exp
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| 	} else {
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| 		mantlo = mant*2 - 1
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| 		explo = exp - 1
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| 	}
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| 	lower := new(decimal)
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| 	lower.Assign(mantlo*2 + 1)
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| 	lower.Shift(explo - int(flt.mantbits) - 1)
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| 
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| 	// The upper and lower bounds are possible outputs only if
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| 	// the original mantissa is even, so that IEEE round-to-even
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| 	// would round to the original mantissa and not the neighbors.
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| 	inclusive := mant%2 == 0
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| 
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| 	// As we walk the digits we want to know whether rounding up would fall
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| 	// within the upper bound. This is tracked by upperdelta:
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| 	//
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| 	// If upperdelta == 0, the digits of d and upper are the same so far.
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| 	//
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| 	// If upperdelta == 1, we saw a difference of 1 between d and upper on a
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| 	// previous digit and subsequently only 9s for d and 0s for upper.
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| 	// (Thus rounding up may fall outside the bound, if it is exclusive.)
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| 	//
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| 	// If upperdelta == 2, then the difference is greater than 1
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| 	// and we know that rounding up falls within the bound.
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| 	var upperdelta uint8
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| 
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| 	// Now we can figure out the minimum number of digits required.
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| 	// Walk along until d has distinguished itself from upper and lower.
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| 	for ui := 0; ; ui++ {
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| 		// lower, d, and upper may have the decimal points at different
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| 		// places. In this case upper is the longest, so we iterate from
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| 		// ui==0 and start li and mi at (possibly) -1.
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| 		mi := ui - upper.dp + d.dp
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| 		if mi >= d.nd {
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| 			break
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| 		}
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| 		li := ui - upper.dp + lower.dp
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| 		l := byte('0') // lower digit
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| 		if li >= 0 && li < lower.nd {
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| 			l = lower.d[li]
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| 		}
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| 		m := byte('0') // middle digit
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| 		if mi >= 0 {
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| 			m = d.d[mi]
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| 		}
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| 		u := byte('0') // upper digit
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| 		if ui < upper.nd {
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| 			u = upper.d[ui]
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| 		}
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| 
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| 		// Okay to round down (truncate) if lower has a different digit
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| 		// or if lower is inclusive and is exactly the result of rounding
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| 		// down (i.e., and we have reached the final digit of lower).
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| 		okdown := l != m || inclusive && li+1 == lower.nd
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| 
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| 		switch {
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| 		case upperdelta == 0 && m+1 < u:
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| 			// Example:
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| 			// m = 12345xxx
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| 			// u = 12347xxx
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| 			upperdelta = 2
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| 		case upperdelta == 0 && m != u:
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| 			// Example:
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| 			// m = 12345xxx
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| 			// u = 12346xxx
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| 			upperdelta = 1
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| 		case upperdelta == 1 && (m != '9' || u != '0'):
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| 			// Example:
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| 			// m = 1234598x
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| 			// u = 1234600x
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| 			upperdelta = 2
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| 		}
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| 		// Okay to round up if upper has a different digit and either upper
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| 		// is inclusive or upper is bigger than the result of rounding up.
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| 		okup := upperdelta > 0 && (inclusive || upperdelta > 1 || ui+1 < upper.nd)
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| 
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| 		// If it's okay to do either, then round to the nearest one.
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| 		// If it's okay to do only one, do it.
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| 		switch {
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| 		case okdown && okup:
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| 			d.Round(mi + 1)
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| 			return
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| 		case okdown:
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| 			d.RoundDown(mi + 1)
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| 			return
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| 		case okup:
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| 			d.RoundUp(mi + 1)
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| 			return
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| 		}
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| 	}
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| }
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