I want to convert uint8 to string but can't figure out how.
package main
import "fmt"
import "strconv"
func main() {
str := "Hello"
fmt.Println(str[1]) // 101
fmt.Println(strconv.Itoa(str[1]))
}
This gives me prog.go:11: cannot use str[1] (type uint8) as type int in function argument
[process exited with non-zero status]
Any idea?
5 Answers
There is a difference between converting it or casting it, consider:
var s uint8 = 10
fmt.Print(string(s))
fmt.Print(strconv.Itoa(int(s)))
The string cast prints '\n' (newline), the string conversion prints "10". The difference becomes clear once you regard the []byte conversion of both variants:
[]byte(string(s)) == [10] // the single character represented by 10
[]byte(strconv.Itoa(int(s))) == [49, 48] // character encoding for '1' and '0'
You can do it even simpler by using casting, this worked for me:
var c uint8
c = 't'
fmt.Printf(string(c))
There are no automatic conversions of basic types in Go expressions. See . A byte (an alias of uint8) or []byte ([]uint8) has to be set to a bool, number or string.
package main
import (
. "fmt"
)
func main() {
b := []byte{'G', 'o'}
c := []interface{}{b[0], float64(b[0]), int(b[0]), rune(b[0]), string(b[0]), Sprintf("%s", b), b[0] != 0}
checkType(c)
}
func checkType(s []interface{}) {
for k, _ := range s {
// uint8 71, float64 71, int 71, int32 71, string G, string Go, bool true
Printf("%T %v\n", s[k], s[k])
}
}
Sprintf("%s", b) can be used to convert []byte{'G', 'o' } to the string "Go". You can convert any int type to a string with Sprintf. See .
But Sprintf uses reflection. See the comment in . Using Itoa (Integer to ASCII) is faster. See @DenysSéguret and . Quotes edited.
use %c
str := "Hello"
fmt.Println(str[1]) // 101
fmt.Printf("%c\n", str[1])
