英文:
Golang - How to turn an exponent and mantissa to a float64 or float32
问题
我有一个库，可以提供浮点数的尾数和指数值，以保证不丢失精度。
尾数：375400000000000
指数：-9
在golang中，我该如何将其转换为float32或float64？
我在这里尝试了一下，但是得到的数字明显不正确：https://go.dev/play/p/8wMxhq11qwS
package main
import (
"fmt"
"math/big"
)
func main() {
mantissa := 375600000000000
exp := int8(-9)
var price big.Float
mant := big.NewFloat(float64(mantissa))
price.SetMantExp(mant, int(exp))
fmt.Println("price: ", price.String())
fmt.Println("mant: ", mantissa)
fmt.Println("exp: ", exp)
}
显然这不起作用，因为big.Float的表示方式是sign × mantissa × 2**exponent，而我需要的是sign × mantissa × 10^exp。
英文:
I have a library that gives floats in mantissa and exponent values for no loss of precision.
Mantissa: 375400000000000
Exponent: -9
How do I turn this into a float32 or float64 in golang?
I gave it a shot here but the number is definitely not coming out correctly: https://go.dev/play/p/8wMxhq11qwS
package main
import (
"fmt"
"math/big"
)
func main() {
mantissa := 375600000000000
exp := int8(-9)
var price big.Float
mant := big.NewFloat(float64(mantissa))
price.SetMantExp(mant, int(exp))
fmt.Println("price: ", price.String())
fmt.Println("mant: ", mantissa)
fmt.Println("exp: ", exp)
}
Apparently this doesn't work because big.Float is represented by sign × mantissa × 2**exponent
when I need sign x mantissa x 10 ^ exp
答案1
得分: 3
我认为最明显的解决方案是使用数学包中的Pow10函数：
package main
import (
"fmt"
"math"
)
func main() {
mantissa := 375600000000000
exp := int8(-9)
price := float64(mantissa) * math.Pow10(int(exp))
fmt.Println("价格：", price)
fmt.Println("尾数：", mantissa)
fmt.Println("指数：", exp)
}
这个函数对于指数的所有可能值都是有效的，但是精度可能会因尾数的值而降低。
英文:
I think the most obvious solution would be to use the Pow10 function in the math package:
package main
import (
"fmt"
"math"
)
func main() {
mantissa := 375600000000000
exp := int8(-9)
price := float64(mantissa) * math.Pow10(int(exp))
fmt.Println("price: ", price)
fmt.Println("mant: ", mantissa)
fmt.Println("exp: ", exp)
}
This function is valid for all possible values of your exponent, but precision may be degraded depending on the value of your mantissa.
答案2
得分: 1
作为一个在Go中经常遇到精度问题的人，math/big包是使用float32或float64的一个很好的替代方案。然而，它并没有提供一个指数函数。一个提供指数函数的包是ericlagergren/decimal：
package main
import (
"fmt"
"github.com/ericlagergren/decimal"
)
func main() {
mantissa := decimal.New(375600000000000, 0)
exp := decimal.New(-9, 0)
ctx := decimal.Context {
Precision: 9, // 设置这个来确定期望的小数位数
}
price := new(decimal.Big)
ctx.Pow(price, decimal.New(10, 0), exp).Mul(price, mantissa)
fmt.Println("price: ", price)
fmt.Println("mant: ", mantissa)
fmt.Println("exp: ", exp)
}
这个包的好处是结果保证符合设置的精度。缺点是你必须显式地设置精度。在这个例子中，我设置了9位数字的精度，因为中间计算会得到一个输出为1e-9的结果，需要9位精度。
英文:
As someone who's had to deal with precision issues in Go quite a bit, the math/big package is an excellent alternative to using float32 or float64. However, it does not offer an exponentiation function. One package that does is ericlagergren/decimal:
package main
import (
"fmt"
"github.com/ericlagergren/decimal"
)
func main() {
mantissa := decimal.New(375600000000000, 0)
exp := decimal.New(-9, 0)
ctx := decimal.Context {
Precision: 9, // Set this to determine the number of expected decimal places
}
price := new(decimal.Big)
ctx.Pow(price, decimal.New(10, 0), exp).Mul(price, mantissa)
fmt.Println("price: ", price)
fmt.Println("mant: ", mantissa)
fmt.Println("exp: ", exp)
}
The upside to this package is that the result is guaranteed to the set precision. The downside is that you have to set that precision explicitly. In this example, I set a precision of 9 digits because an interim calculation would result in an output of 1e-9, which requires 9 digits of precision.
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