I would like to format a big int as a string with leading zeros. I'm looking for an example similar to this, but with Big:
I'm looking at source here.
But when I call:
m := big.NewInt(99999999999999)
fmt.Println(m.Format("%010000000000000000000","d"))
I see:
prog.go:10:22: m.Format("%010000000000000000000", "d") used as value
prog.go:10:23: cannot use "%010000000000000000000" (type string) as type fmt.State in argument to m.Format:
string does not implement fmt.State (missing Flag method)
prog.go:10:48: cannot use "d" (type string) as type rune in argument to m.Format
(I understand normally I can use m.String(), but zero padding seems to complicate this, so I'm looking specifically for some help on the Format method.)
Here's my playground link.
You can simply use fmt.Sprintf(...) with the "%020s" directive (where 20 is whatever total length you want). The s verb will use the natural string format of the big int and the 020 modifier will create a string with total length of (at least) 20 with zero padding (instead of whitespace).
For example (Go Playground):
m := big.NewInt(99999999999999)
s := fmt.Sprintf("%020s", m)
fmt.Println(s)
// 00000099999999999999
The Int.Format() is not for you to call manually (although you could), but it is to implement fmt.Formatter so the fmt package will support formatting big.Int values out-of-the box.
See this example:
m := big.NewInt(99)
fmt.Printf("%06d
", m)
if _, ok := m.SetString("1234567890123456789012345678901234567890", 10); !ok {
panic("big")
}
fmt.Printf("%060d
", m)
Outputs (try it on the Go Playground):
000099
000000000000000000001234567890123456789012345678901234567890
This is the simplest, so use this. Manually creating an fmt.Formatter gives you more control, but is also harder to do it. Unless this is a performance critical part of your app, just use the above solution.