CODEWITHSUNDEEP
go
qed
qed

Reputation: 23134

Strange pow implementation in golang

I just came across the Pow implementation in golang:

func Pow(x, y float64) float64 {
    // ...
    case x == 0:
        switch {
        case y < 0:
            if isOddInt(y) {
                return Copysign(Inf(1), x)
            }
            return Inf(1)
        case y > 0:
            if isOddInt(y) {
                return x
            }
            return 0
        }
    //...
}

Isn't the case y > 0 part over complicated? I would just return 0. Or did I miss something?

Upvotes: 4

Views: 546

Answers (2)

ymonad
ymonad

Reputation: 12100

there are two types of zero's, +0 and -0. return value of Pow(-0,1) should be -0 not +0

to create -0 in golang, use math.Copysign.

x := math.Copysign(0, -1)
if x == 0 {
    fmt.Println("x is zero")
}
fmt.Println("x ** 3 is", math.Pow(x, 3))

the output of above code is

x is zero
x ** 3 is -0

you can check it in Go Playground

why we have to distinguish +0 and -0, see: https://softwareengineering.stackexchange.com/questions/280648/why-is-negative-zero-important

Upvotes: 4

Mike Vella
Mike Vella

Reputation: 10575

The answer is in the inline documentation for the function, in the case case y > 0 then the function output is as follows:

Pow(±0, y) = ±0 for y an odd integer > 0
Pow(±0, y) = +0 for finite y > 0 and not an odd integer

so the function will only return 0 (+0) like you say in the case that x=0

Upvotes: 1

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