In Golang, it seems that when a float64 var first convert to float32 then convert float64, it's value will change.
a := -8888.95
fmt.Println(a) // -8888.95
fmt.Println(float32(a)) // -8888.95
fmt.Println(float64(float32(a))) // -8888.9501953125
How can I make it unchanging
The way you have described the problem is perhaps misleading.
The precision is not lost "when converting float32 to float64"; rather, it is lost when converting from float64 to float32.
So how can you avoid losing precision when converting from float64 to float32? You can't. This task is impossible, and it's quite easy to see the reason why:
float64 has twice as many bits as float32
multiple different float64 values will map to the same float32 value due to the pigeonhole principle
the conversion is therefore not reversible.
package main
import (
"fmt"
)
func main() {
a := -8888.95
fmt.Printf("%.20f\n", a)
fmt.Printf("%.20f\n", float32(a))
fmt.Printf("%.20f\n", float64(float32(a)))
}
Adjusting your program to show a more precise output of each value, you'll see exactly where the precision is lost:
-8888.95000000000072759576
-8888.95019531250000000000
-8888.95019531250000000000
That is, after the float32 conversion (as is expected).
It's also worth noting that neither float64 nor float32 can represent your value -8888.95 exactly. If you convert this number to a fraction, you will get -177779/20. Notice the denominator, 20. The prime factorization of 20 is 2 * 2 * 5.
If you apply this process to a number and the prime factorization of the denominator contains any factors which are NOT 2, then you can rest assured that this number is definitely not representable exactly in binary floating point form. You may discover that the probability of any number passing this test is quite low.
Related
I am aware of the int() or int64() typecast functions, the Go provides.
Is there a safe way to cast float64 to int64 (with truncation,) given the largest representable 64-bit floating point number math.MaxFloat64 is much larger than math.MaxInt64? A way, that triggers an overflow error?
If you are converting (not casting or typecasting) from a float64 to an int64, then you can compare the float value to the closest double precision integer values of math.MaxInt64 and math.MinInt64. These numbers cannot be exactly stored in a float64 value, however they will be stored as the next integer value above or below those:
float64(math.MaxInt64) // 9223372036854775808
float64(math.MinInt64) // -9223372036854775808
So as long as your float64 value is between these two values, you know it can be converted to an int64 without overflow.
Note however that if you are converting in the other direction, you cannot guarantee that a float64 can accurately represent all int64 values up to MaxInt64, since a double precision float point value only holds 52 bits of precision. In that case you must check that the int64 value is between -1<<53 and 1<<53.
In either case you can handle larger values with the math/big package.
I need to find the square root of a big.Rat. Is there a way to do it without losing (already existing) accuracy?
For example, I could convert the numerator and denominator into floats, get the square root, and then convert it back...
func ratSquareRoot(num *big.Rat) *big.Rat {
f, exact := num.Float64() //Yuck! Floats!
squareRoot := math.Sqrt(f)
var accuracy int64 = 10 ^ 15 //Significant digits of precision for float64
return big.NewRat(int64(squareRoot*float64(accuracy)), accuracy)
// ^ This is now totally worthless. And also probably not simplified very well.
}
...but that would eliminate all of the accuracy of using a rational. Is there a better way of doing this?
The big.Float type has a .Sqrt(x) operation, and handles defining explicitly the precision you aim for. I'd try to use that and convert the result back to a Rat with the same operations in your question, only manipulating big.Int values.
r := big.NewRat(1, 3)
var x big.Float
x.SetPrec(30) // I didn't figure out the 'Prec' part correctly, read the docs more carefully than I did and experiement
x.SetRat(r)
var s big.Float
s.SetPrec(15)
s.Sqrt(&x)
r, _ = s.Rat(nil)
fmt.Println(x.String(), s.String())
fmt.Println(r.String(), float64(18919)/float64(32768))
playground
I've been reading this post on constants in Go, and I'm trying to understand how they are stored and used in memory. You can perform operations on very large constants in Go, and as long as the result fits in memory, you can coerce that result to a type. For example, this code prints 10, as you would expect:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
How does this work under the hood? At some point, Go has to store 1e1000 and 1e999 in memory, in order to perform operations on them. So how are constants stored, and how does Go perform arithmetic on them?
Short summary (TL;DR) is at the end of the answer.
Untyped arbitrary-precision constants don't live at runtime, constants live only at compile time (during the compilation). That being said, Go does not have to represent constants with arbitrary precision at runtime, only when compiling your application.
Why? Because constants do not get compiled into the executable binaries. They don't have to be. Let's take your example:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
There is a constant Huge in the source code (and will be in the package object), but it won't appear in your executable. Instead a function call to fmt.Println() will be recorded with a value passed to it, whose type will be float64. So in the executable only a float64 value being 10.0 will be recorded. There is no sign of any number being 1e1000 in the executable.
This float64 type is derived from the default type of the untyped constant Huge. 1e1000 is a floating-point literal. To verify it:
const Huge = 1e1000
x := Huge / 1e999
fmt.Printf("%T", x) // Prints float64
Back to the arbitrary precision:
Spec: Constants:
Numeric constants represent exact values of arbitrary precision and do not overflow.
So constants represent exact values of arbitrary precision. As we saw, there is no need to represent constants with arbitrary precision at runtime, but the compiler still has to do something at compile time. And it does!
Obviously "infinite" precision cannot be dealt with. But there is no need, as the source code itself is not "infinite" (size of the source is finite). Still, it's not practical to allow truly arbitrary precision. So the spec gives some freedom to compilers regarding to this:
Implementation restriction: Although numeric constants have arbitrary precision in the language, a compiler may implement them using an internal representation with limited precision. That said, every implementation must:
Represent integer constants with at least 256 bits.
Represent floating-point constants, including the parts of a complex constant, with a mantissa of at least 256 bits and a signed exponent of at least 32 bits.
Give an error if unable to represent an integer constant precisely.
Give an error if unable to represent a floating-point or complex constant due to overflow.
Round to the nearest representable constant if unable to represent a floating-point or complex constant due to limits on precision.
These requirements apply both to literal constants and to the result of evaluating constant expressions.
However, also note that when all the above said, the standard package provides you the means to still represent and work with values (constants) with "arbitrary" precision, see package go/constant. You may look into its source to get an idea how it's implemented.
Implementation is in go/constant/value.go. Types representing such values:
// A Value represents the value of a Go constant.
type Value interface {
// Kind returns the value kind.
Kind() Kind
// String returns a short, human-readable form of the value.
// For numeric values, the result may be an approximation;
// for String values the result may be a shortened string.
// Use ExactString for a string representing a value exactly.
String() string
// ExactString returns an exact, printable form of the value.
ExactString() string
// Prevent external implementations.
implementsValue()
}
type (
unknownVal struct{}
boolVal bool
stringVal string
int64Val int64 // Int values representable as an int64
intVal struct{ val *big.Int } // Int values not representable as an int64
ratVal struct{ val *big.Rat } // Float values representable as a fraction
floatVal struct{ val *big.Float } // Float values not representable as a fraction
complexVal struct{ re, im Value }
)
As you can see, the math/big package is used to represent untyped arbitrary precision values. big.Int is for example (from math/big/int.go):
// An Int represents a signed multi-precision integer.
// The zero value for an Int represents the value 0.
type Int struct {
neg bool // sign
abs nat // absolute value of the integer
}
Where nat is (from math/big/nat.go):
// An unsigned integer x of the form
//
// x = x[n-1]*_B^(n-1) + x[n-2]*_B^(n-2) + ... + x[1]*_B + x[0]
//
// with 0 <= x[i] < _B and 0 <= i < n is stored in a slice of length n,
// with the digits x[i] as the slice elements.
//
// A number is normalized if the slice contains no leading 0 digits.
// During arithmetic operations, denormalized values may occur but are
// always normalized before returning the final result. The normalized
// representation of 0 is the empty or nil slice (length = 0).
//
type nat []Word
And finally Word is (from math/big/arith.go)
// A Word represents a single digit of a multi-precision unsigned integer.
type Word uintptr
Summary
At runtime: predefined types provide limited precision, but you can "mimic" arbitrary precision with certain packages, such as math/big and go/constant. At compile time: constants seemingly provide arbitrary precision, but in reality a compiler may not live up to this (doesn't have to); but still the spec provides minimal precision for constants that all compiler must support, e.g. integer constants must be represented with at least 256 bits which is 32 bytes (compared to int64 which is "only" 8 bytes).
When an executable binary is created, results of constant expressions (with arbitrary precision) have to be converted and represented with values of finite precision types – which may not be possible and thus may result in compile-time errors. Note that only results –not intermediate operands– have to be converted to finite precision, constant operations are carried out with arbitrary precision.
How this arbitrary or enhanced precision is implemented is not defined by the spec, math/big for example stores "digits" of the number in a slice (where digits is not a digit of the base 10 representation, but "digit" is an uintptr which is like base 4294967295 representation on 32-bit architectures, and even bigger on 64-bit architectures).
Go constants are not allocated to memory. They are used in context by the compiler. The blog post you refer to gives the example of Pi:
Pi = 3.14159265358979323846264338327950288419716939937510582097494459
If you assign Pi to a float32 it will lose precision to fit, but if you assign it to a float64, it will lose less precision, but the compiler will determine what type to use.
If I run the following piece of Go code:
fmt.Println(float32(0.1) + float32(0.2))
fmt.Println(float64(0.1) + float64(0.2))
the output is:
0.3
0.30000000000000004
It appears the result of the float32 sum is more exact than the result of the float64 sum, why? I thought that float64 is always more precise than float32. How do I decide which one to pick to have the most accurate result?
It isn't. fmt.Println is just making it look more precise. Println uses %g for floating point and complex numbers. The docs say...
The default precision for... %g it is the smallest number of digits necessary to identify the value uniquely.
0.3 is sufficient to identify a float32. But float64 being much more precise needs more digits.
We can use fmt.Printf and %0.20g to force both numbers to display the same precision.
f32 := float32(0.1) + float32(0.2)
f64 := float64(0.1) + float64(0.2)
fmt.Printf("%0.20g\n", f32)
fmt.Printf("%0.20g\n", f64)
0.30000001192092895508
0.30000000000000004441
float64 is more precise. Neither are exact as that is the nature of floating point numbers.
We can use strconv.FormatFloat to see what these numbers really are.
fmt.Println(strconv.FormatFloat(float64(f32), 'b', -1, 32))
fmt.Println(strconv.FormatFloat(f64, 'b', -1, 64))
10066330p-25
5404319552844596p-54
That is 10066330 * 2^-25 and 5404319552844596 * 2^-54.
I've been reading this post on constants in Go, and I'm trying to understand how they are stored and used in memory. You can perform operations on very large constants in Go, and as long as the result fits in memory, you can coerce that result to a type. For example, this code prints 10, as you would expect:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
How does this work under the hood? At some point, Go has to store 1e1000 and 1e999 in memory, in order to perform operations on them. So how are constants stored, and how does Go perform arithmetic on them?
Short summary (TL;DR) is at the end of the answer.
Untyped arbitrary-precision constants don't live at runtime, constants live only at compile time (during the compilation). That being said, Go does not have to represent constants with arbitrary precision at runtime, only when compiling your application.
Why? Because constants do not get compiled into the executable binaries. They don't have to be. Let's take your example:
const Huge = 1e1000
fmt.Println(Huge / 1e999)
There is a constant Huge in the source code (and will be in the package object), but it won't appear in your executable. Instead a function call to fmt.Println() will be recorded with a value passed to it, whose type will be float64. So in the executable only a float64 value being 10.0 will be recorded. There is no sign of any number being 1e1000 in the executable.
This float64 type is derived from the default type of the untyped constant Huge. 1e1000 is a floating-point literal. To verify it:
const Huge = 1e1000
x := Huge / 1e999
fmt.Printf("%T", x) // Prints float64
Back to the arbitrary precision:
Spec: Constants:
Numeric constants represent exact values of arbitrary precision and do not overflow.
So constants represent exact values of arbitrary precision. As we saw, there is no need to represent constants with arbitrary precision at runtime, but the compiler still has to do something at compile time. And it does!
Obviously "infinite" precision cannot be dealt with. But there is no need, as the source code itself is not "infinite" (size of the source is finite). Still, it's not practical to allow truly arbitrary precision. So the spec gives some freedom to compilers regarding to this:
Implementation restriction: Although numeric constants have arbitrary precision in the language, a compiler may implement them using an internal representation with limited precision. That said, every implementation must:
Represent integer constants with at least 256 bits.
Represent floating-point constants, including the parts of a complex constant, with a mantissa of at least 256 bits and a signed exponent of at least 32 bits.
Give an error if unable to represent an integer constant precisely.
Give an error if unable to represent a floating-point or complex constant due to overflow.
Round to the nearest representable constant if unable to represent a floating-point or complex constant due to limits on precision.
These requirements apply both to literal constants and to the result of evaluating constant expressions.
However, also note that when all the above said, the standard package provides you the means to still represent and work with values (constants) with "arbitrary" precision, see package go/constant. You may look into its source to get an idea how it's implemented.
Implementation is in go/constant/value.go. Types representing such values:
// A Value represents the value of a Go constant.
type Value interface {
// Kind returns the value kind.
Kind() Kind
// String returns a short, human-readable form of the value.
// For numeric values, the result may be an approximation;
// for String values the result may be a shortened string.
// Use ExactString for a string representing a value exactly.
String() string
// ExactString returns an exact, printable form of the value.
ExactString() string
// Prevent external implementations.
implementsValue()
}
type (
unknownVal struct{}
boolVal bool
stringVal string
int64Val int64 // Int values representable as an int64
intVal struct{ val *big.Int } // Int values not representable as an int64
ratVal struct{ val *big.Rat } // Float values representable as a fraction
floatVal struct{ val *big.Float } // Float values not representable as a fraction
complexVal struct{ re, im Value }
)
As you can see, the math/big package is used to represent untyped arbitrary precision values. big.Int is for example (from math/big/int.go):
// An Int represents a signed multi-precision integer.
// The zero value for an Int represents the value 0.
type Int struct {
neg bool // sign
abs nat // absolute value of the integer
}
Where nat is (from math/big/nat.go):
// An unsigned integer x of the form
//
// x = x[n-1]*_B^(n-1) + x[n-2]*_B^(n-2) + ... + x[1]*_B + x[0]
//
// with 0 <= x[i] < _B and 0 <= i < n is stored in a slice of length n,
// with the digits x[i] as the slice elements.
//
// A number is normalized if the slice contains no leading 0 digits.
// During arithmetic operations, denormalized values may occur but are
// always normalized before returning the final result. The normalized
// representation of 0 is the empty or nil slice (length = 0).
//
type nat []Word
And finally Word is (from math/big/arith.go)
// A Word represents a single digit of a multi-precision unsigned integer.
type Word uintptr
Summary
At runtime: predefined types provide limited precision, but you can "mimic" arbitrary precision with certain packages, such as math/big and go/constant. At compile time: constants seemingly provide arbitrary precision, but in reality a compiler may not live up to this (doesn't have to); but still the spec provides minimal precision for constants that all compiler must support, e.g. integer constants must be represented with at least 256 bits which is 32 bytes (compared to int64 which is "only" 8 bytes).
When an executable binary is created, results of constant expressions (with arbitrary precision) have to be converted and represented with values of finite precision types – which may not be possible and thus may result in compile-time errors. Note that only results –not intermediate operands– have to be converted to finite precision, constant operations are carried out with arbitrary precision.
How this arbitrary or enhanced precision is implemented is not defined by the spec, math/big for example stores "digits" of the number in a slice (where digits is not a digit of the base 10 representation, but "digit" is an uintptr which is like base 4294967295 representation on 32-bit architectures, and even bigger on 64-bit architectures).
Go constants are not allocated to memory. They are used in context by the compiler. The blog post you refer to gives the example of Pi:
Pi = 3.14159265358979323846264338327950288419716939937510582097494459
If you assign Pi to a float32 it will lose precision to fit, but if you assign it to a float64, it will lose less precision, but the compiler will determine what type to use.