How can I convert from int to hex

Question

I want to convert from int to hex in Golang. In strconv, there is a method that converts strings to hex. Is there a similar method to get a hex string from an int?

Answer 1

If formatting some bytes, hex needs a 2 digits representation, with leading 0.

For exemple: 1 => '01', 15 => '0f', etc.

It is possible to force Sprintf to respect this :

h:= fmt.Sprintf("%02x", 14)
fmt.Println(h) // 0e
h2:= fmt.Sprintf("%02x", 231)
fmt.Println(h2) // e7

The pattern "%02x" means:

• '0' force using zeros
• '2' set the output size as two charactes
• 'x' to convert in hexadecimal

Answer 2

Since hex is a Integer literal, you can ask the fmt package for a string representation of that integer, using fmt.Sprintf(), and the %x or %X format.
See playground

i := 255
h := fmt.Sprintf("%x", i)
fmt.Printf("Hex conv of '%d' is '%s'\n", i, h)
h = fmt.Sprintf("%X", i)
fmt.Printf("HEX conv of '%d' is '%s'\n", i, h)

Output:

Hex conv of '255' is 'ff'
HEX conv of '255' is 'FF'

Answer 3

E.g. if its uint32, you can convert it to HEX as seen below =>

    var p uint32
    p = 4278190335
    r := p >> 24 & 0xFF
    g := p >> 16 & 0xFF
    b := p >> 8 & 0xFF
    fmt.Println(r, g, b)//255 0 0

DEMO

you can also check this online tool for ref. https://cryptii.com/pipes/integer-encoder

Answer 4

"Hex" isn't a real thing. You can use a hexadecimal representation of a number, but there's no difference between 0xFF and 255. More info on that can be found in the docs which point out you can use 0xff to define an integer constant 255! As you mention, if you're trying to find the hexadecimal representation of an integer you could use strconv

package main

import (
    "fmt"
    "strconv"
)

func main() {
    fmt.Println(strconv.FormatInt(255, 16))
    // gives "ff"
}

Try it in the playground

Answer 5

i := 4357640193405743614
h := fmt.Sprintf("%016x",i)
fmt.Printf("Decimal: %d,\nHexa: %s", i, h)

# Result
Decimal..: 4357640193405743614,
Hexa.....: 3c7972ab0ae9f1fe

Playground: https://play.golang.org/p/ndlMyBdQjmT

Answer 6

Sprintf is more versatile but FormatInt is faster. Choose what is better for you

func Benchmark_sprintf(b *testing.B) { // 83.8 ns/op
    for n := 0; n < b.N; n++ {
        _ = fmt.Sprintf("%x", n)
    }
}
func Benchmark_formatint(b *testing.B) { // 28.5 ns/op
    bn := int64(b.N)
    for n := int64(0); n < bn; n++ {
        _ = strconv.FormatInt(n, 16)
    }
}