How to index characters in a Golang string?

Question

How to get an \"E\" output rather than 69?

package main

import \"fmt\"

func main() {
    fmt.Print(\"HELLO\"[1])
}

Does Golang have funct

Answer 1

Try this to get the charecters by their index

package main

import (
      "fmt"
      "strings"
)

func main() {
   str := strings.Split("HELLO","")
    fmt.Print(str[1])
}

Answer 2

Go doesn't really have a character type as such. byte is often used for ASCII characters, and rune is used for Unicode characters, but they are both just aliases for integer types (uint8 and int32). So if you want to force them to be printed as characters instead of numbers, you need to use Printf("%c", x). The %c format specification works for any integer type.

Answer 3

Interpreted string literals are character sequences between double quotes "" using the (possibly multi-byte) UTF-8 encoding of individual characters. In UTF-8, ASCII characters are single-byte corresponding to the first 128 Unicode characters. Strings behave like slices of bytes. A rune is an integer value identifying a Unicode code point. Therefore,

package main

import "fmt"

func main() {
    fmt.Println(string("Hello"[1]))              // ASCII only
    fmt.Println(string([]rune("Hello, 世界")[1])) // UTF-8
    fmt.Println(string([]rune("Hello, 世界")[8])) // UTF-8
}

Output:

e
e
界

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Read:

Go Programming Language Specification section on Conversions.

The Go Blog: Strings, bytes, runes and characters in Go