CODEWITHSUNDEEP
go
Chris Bunch
Chris Bunch

Reputation: 89893

Convert a float64 to an int in Go

How does one convert a float64 to an int in Go? I know the strconv package can be used to convert anything to or from a string, but not between data types where one isn't a string. I know I can use fmt.Sprintf to convert anything to a string, and then strconv it to the data type I need, but this extra conversion seems a bit clumsy - is there a better way to do this?

Upvotes: 227

Views: 389989

Answers (5)

scaszoo
scaszoo

Reputation: 311

Correct rounding is likely desired.

Therefore math.Round() is your quick(!) friend. Approaches with fmt.Sprintf and strconv.Atois() were 2 orders of magnitude slower according to my tests with a matrix of float64 values that were intended to become correctly rounded int values.

package main
import (
    "fmt"
    "math"
)
func main() {
    var x float64 = 5.51
    var y float64 = 5.50
    var z float64 = 5.49
    fmt.Println(int(math.Round(x)))  // outputs "6"
    fmt.Println(int(math.Round(y)))  // outputs "6"
    fmt.Println(int(math.Round(z)))  // outputs "5"
}

math.Round() does return a float64 value but with int() applied afterwards, I couldn't find any mismatches so far.

Upvotes: 18

Codemaker2015
Codemaker2015

Reputation: 15679

You can use int() function to convert float64 type data to an int. Similarly you can use float64()

Example:

func check(n int) bool { 
    // count the number of digits 
    var l int = countDigit(n)
    var dup int = n 
    var sum int = 0 

    // calculates the sum of digits 
    // raised to power 
    for dup > 0 { 
        **sum += int(math.Pow(float64(dup % 10), float64(l)))** 
        dup /= 10 
    } 

    return n == sum
}

Upvotes: 4

Peanuts
Peanuts

Reputation: 352

If its simply from float64 to int, this should work

package main

import (
    "fmt"
)

func main() {
    nf := []float64{-1.9999, -2.0001, -2.0, 0, 1.9999, 2.0001, 2.0}

    //round
    fmt.Printf("Round : ")
    for _, f := range nf {
        fmt.Printf("%d ", round(f))
    }
    fmt.Printf("\n")

    //rounddown ie. math.floor
    fmt.Printf("RoundD: ")
    for _, f := range nf {
        fmt.Printf("%d ", roundD(f))
    }
    fmt.Printf("\n")

    //roundup ie. math.ceil
    fmt.Printf("RoundU: ")
    for _, f := range nf {
        fmt.Printf("%d ", roundU(f)) 
    }
    fmt.Printf("\n")

}

func roundU(val float64) int {
    if val > 0 { return int(val+1.0) }
    return int(val)
}

func roundD(val float64) int {
    if val < 0 { return int(val-1.0) }
    return int(val)
}

func round(val float64) int {
    if val < 0 { return int(val-0.5) }
    return int(val+0.5)
}

Outputs:

Round : -2 -2 -2 0 2 2 2 
RoundD: -2 -3 -3 0 1 2 2 
RoundU: -1 -2 -2 0 2 3 3

Here's the code in the playground - https://play.golang.org/p/HmFfM6Grqh

Upvotes: 6

David Dooling
David Dooling

Reputation: 864

Simply casting to an int truncates the float, which if your system internally represent 2.0 as 1.9999999999, you will not get what you expect. The various printf conversions deal with this and properly round the number when converting. So to get a more accurate value, the conversion is even more complicated than you might first expect:

package main

import (
    "fmt"
    "strconv"
)

func main() {
    floats := []float64{1.9999, 2.0001, 2.0}
    for _, f := range floats {
        t := int(f)
        s := fmt.Sprintf("%.0f", f)
        if i, err := strconv.Atoi(s); err == nil {
            fmt.Println(f, t, i)
        } else {
            fmt.Println(f, t, err)
        }
    }
}

Code on Go Playground

Upvotes: 21

David Grayson
David Grayson

Reputation: 87506

package main
import "fmt"
func main() {
  var x float64 = 5.7
  var y int = int(x)
  fmt.Println(y)  // outputs "5"
}

Upvotes: 333

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