float64 的值可以支持多小？

Question

I get data from an api request and perform a calculation based on the value.

Let say my bank account or crypto account have a balance of 301.38481999999999

I want to withdraw all of the money, but the float64 variable in golang automatically round up the variable to 301.38482 which caused the withdrawal operation to fail because I do not have that much money in my account.

package main

import(
    "log"
)

func main(){

    var myvar float64

    myvar = 301.38481999999999
    log.Println(myvar)
}

https://play.golang.org/p/BXk9fcVJZVn

It show

301.38482

How can I get the exact balance with exact number so that I can withdraw all of it?

Answer 1

You're confusing the output you see with the actual value you have. If you check the docs of fmt :

For floating-point values, width sets the minimum width of the field and precision sets the number of places after the decimal, if appropriate, except that for %g/%G precision sets the maximum number of significant digits (trailing zeros are removed). For example, given 12.345 the format %6.3f prints 12.345 while %.3g prints 12.3. The default precision for %e, %f and %#g is 6; for %g it is the smallest number of digits necessary to identify the value uniquely.

fmt.Println always uses the default precision for whatever value you pass in. To see more precision, specify more precision:

var myvar float64
myvar = 301.38481999999999
log.Printf("%.20f", myvar)

// 301.38481999999999061401

https://play.golang.org/p/T4iyQcHY4qi

Regardless of what's printed, all internal arithmetic will be done with the full precision of the value. If you need more precision than float64 provides, see the math/big package.