Golang : Convert long hexadecimal with strconv.ParseUint example
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
See also : Golang : Convert integer to binary, octal, hexadecimal and back to integer
By Adam Ng
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