Strange loss of precision multiplying big.Float

If you parse a string into a big.Float like f.SetString("0.001"), then multiply it, I'm seeing a loss of precision. If I use f.SetFloat64(0.001), I don't lose precision. Even doing a strconv.ParseFloat("0.001", 64), then calling f.SetFloat() works.

f.SetString("0.001")

f.SetFloat64(0.001)

strconv.ParseFloat("0.001", 64)

f.SetFloat()

Full example of what I'm seeing here:

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

Expanded from this question: https://stackoverflow.com/a/47546136/105562

1 Answer
1

The difference in output is due to imprecise representation of base 10 floating point numbers in float64 (IEEE-754 format) and the default precision and rounding of big.Float.

float64

big.Float

See this simple code to verify:

fmt.Printf("%.30fn", 0.001) f, ok := new(big.Float).SetString("0.001") fmt.Println(f.Prec(), ok)

Output of the above (try it on the Go Playground):

0.001000000000000000020816681712 64 true

So what we see is that the float64 value 0.001 is not exactly 0.001, and the default precision of big.Float is 64.

float64

0.001

big.Float

If you increase the precision of the number you set via a string value, you will see the same output:

string

s := "0.001" f := new(big.Float) f.SetPrec(100) f.SetString(s) fmt.Println(s) fmt.Println(BigFloatToBigInt(f))

Now output will also be the same (try it on the Go Playground):

0.001 1000000000000000

Thanks, that works, and I had actually tried that, but I used SetPrec(128) and that doesn't work. Turns out 128 and 64 don't work, work but everything around it does... Any idea why that is? See results here: play.golang.org/p/KLnPY8cqKpe
– Travis Reeder
Sep 4 '18 at 15:01

SetPrec(128)

@TravisReeder I assume it is due to the fact that 0.001 cannot be represented precisely, and when using finite bits, the implementation rounds 0.001 to the closest number representable using the given precision. Sometimes it might mean to round up, sometimes to round down.
– icza
Sep 4 '18 at 15:26

0.001

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