How to convert int[] to uint8[] - type-conversion

Question

SO, I need your help. I couldn’t find anything on that topic. Golang is a freshly baked language so it’s quite hard to find answers quick for a newcomers like me.

Answer 1

The predeclared Go int type size is implementation-specific, either 32 or 64 bits (Numeric types).
Here's an example of converting big-endian ints to bytes (uint8s).
package main
import (
"encoding/binary"
"fmt"
"reflect"
)
func IntsToBytesBE(i []int) []byte {
intSize := int(reflect.TypeOf(i).Elem().Size())
b := make([]byte, intSize*len(i))
for n, s := range i {
switch intSize {
case 64 / 8:
binary.BigEndian.PutUint64(b[intSize*n:], uint64(s))
case 32 / 8:
binary.BigEndian.PutUint32(b[intSize*n:], uint32(s))
default:
panic("unreachable")
}
}
return b
}
func main() {
i := []int{0, 1, 2, 3}
fmt.Println("int size:", int(reflect.TypeOf(i[0]).Size()), "bytes")
fmt.Println("ints:", i)
fmt.Println("bytes:", IntsToBytesBE(i))
}
Output:
int size: 4 bytes
ints: [0 1 2 3]
bytes: [0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 3]
or
int size: 8 bytes
ints: [0 1 2 3]
bytes: [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 3]