Converting string to byte in go

Question

Basically what I'm trying to do is convert a string in the hexidecimal format to a byte and append the byte to a byte slice.

I've tried:

func main() {
    bytes := []byte{0xfc}
    string := "0xe8"
    bytes = append(bytes, string...)
    fmt.Printf("%s", bytes)
}

output:

�0xe8

I know I could just declare a byte variable and append the byte. I need to convert the string into a byte.

Expected output:

��

Tried: "b, err := strconv.ParseInt(string, 10, 64)" outputed error: "strconv.ParseInt: parsing "0xe8": invalid syntax", Could you provide an example? — Justin Galloway
– Justin Galloway, Commented Dec 7, 2019 at 10:47

Alexey Bril · Accepted Answer · 2019-12-07 17:46:52Z

2

This is work fo me.

func main() {
    bytes := []byte{0xfc}
    str := "0xe8"
    pc, _ := strconv.ParseUint(str, 0, 64)

    bytes = append(bytes, uint8(pc))
    fmt.Printf("%s", bytes)
}

Output:

��

edited Dec 7, 2019 at 17:46

answered Dec 7, 2019 at 11:51

Alexey Bril

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3 Comments

Alexey Bril Over a year ago

@Flimzy ok, i changed my post.

mkopriva Over a year ago

@AlexeyBril note that there is no need for the prefix trimming, if you pass in 0 as the base ParseInt and ParseUint will infer the base from the prefix. e.g. strconv.ParseUint("0xe8", 0, 64). From the docs: "If base == 0, the base is implied by the string's prefix: base 2 for "0b", base 8 for "0" or "0o", base 16 for "0x", and base 10 otherwise."

Alexey Bril Over a year ago

@mkopriva Yes, I agree, this option will be more optimal. Fixed implementation

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