Golang : Convert long hexadecimal with strconv.ParseUint example



Tags : golang parseuint uint64 long-hexadecimal integer

Just an example on how to convert a long hexadecimal input string into integer. Most hexadecimal can be handled easily by strconv.ParseInt(hexString, 16, 64) function, but there are times when you will encounter a long hexadecimal and ParseInt() function will not be able to process the long hexadecimal. It will return either -1 or some weird result.

In order to cater for the long hexadecimal, you need to use strconv.ParseUint() instead of ParseInt().

Yup, just use uint64 type for returning the result.

Here you go!

 package main

 import (
  "fmt"
  "strconv"
  "strings"
 )

 func hex2int(hexStr string) uint64 {
  // remove 0x suffix if found in the input string
  cleaned := strings.Replace(hexStr, "0x", "", -1)

  // base 16 for hexadecimal
  result, _ := strconv.ParseUint(cleaned, 16, 64)
  return uint64(result)
 }

 func main() {
  hexString := "0x75bcd15"
  fmt.Println(hex2int(hexString))

  longHexString := "0xffb969d28651e43c"
  fmt.Println(hex2int(longHexString))

  longHexString = "0xFFB969D28651E43C"
  fmt.Println(hex2int(longHexString))

 }

Output:

123456789

18426875703280657468

18426875703280657468

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  See also : Golang : Convert integer to binary, octal, hexadecimal and back to integer



Tags : golang parseuint uint64 long-hexadecimal integer

By Adam Ng

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