转到：使用乘数将float64转换为int

Question

I want to convert a float64 number, let's say it 1.003 to 1003 (integer type). My implementation is simply multiply the float64 with 1000 and cast it to int.

package main

import "fmt"


func main() {
  var f float64 = 1.003
  fmt.Println(int(f * 1000))
}

But when I run that code, what I got is 1002 not 1003. Because Go automatically stores 1.003 as 1.002999... in the variable. What is the correct approach to do this kind of operation on Golang?

Answer 1

Go spec: Conversions:

Conversions between numeric types

When converting a floating-point number to an integer, the fraction is discarded (truncation towards zero).

So basically when you convert a floating-point number to an integer, only the integer part is kept.

If you just want to avoid errors arising from representing with finite bits, just add 0.5 to the number before converting it to int. No external libraries or function calls (from standard library) required.

Since float -> int conversion is not rounding but keeping the integer part, this will give you the desired result. Taking into consideration both the possible smaller and greater representation:

1002.9999 + 0.5 = 1003.4999;     integer part: 1003
1003.0001 + 0.5 = 1003.5001;     integer part: 1003

So simply just write:

var f float64 = 1.003
fmt.Println(int(f * 1000 + 0.5))

To wrap this into a function:

func toint(f float64) int {
    return int(f + 0.5)
}

// Using it:
fmt.Println(toint(f * 1000))

Try them on the Go Playground.

Note:

Be careful when you apply this in case of negative numbers! For example if you have a value of -1.003, then you probably want the result to be -1003. But if you add 0.5 to it:

-1002.9999 + 0.5 = -1002.4999;     integer part: -1002
-1003.0001 + 0.5 = -1002.5001;     integer part: -1002

So if you have negative numbers, you have to either:

• subtract 0.5 instead of adding it
• or add 0.5 but subtract 1 from the result

Incorporating this into our helper function:

func toint(f float64) int {
    if f < 0 {
        return int(f - 0.5)
    }
    return int(f + 0.5)
}