How to convert [4]uint8 into uint32 in Go?
go, type-conversion
Solution
package main
import (
"encoding/binary"
"fmt"
)
func main() {
u8 := []uint8{0, 1, 2, 3}
u32LE := binary.LittleEndian.Uint32(u8)
fmt.Println("little-endian:", u8, "to", u32LE)
u32BE := binary.BigEndian.Uint32(u8)
fmt.Println("big-endian: ", u8, "to", u32BE)
}
Output:
little-endian: [0 1 2 3] to 50462976
big-endian: [0 1 2 3] to 66051
The Go binary package functions are implemented as a series of shifts.
func (littleEndian) Uint32(b []byte) uint32 {
return uint32(b[0]) | uint32(b[1])<<8 | uint32(b[2])<<16 | uint32(b[3])<<24
}
func (bigEndian) Uint32(b []byte) uint32 {
return uint32(b[3]) | uint32(b[2])<<8 | uint32(b[1])<<16 | uint32(b[0])<<24
}
Problem
how to convert go's type from uint8 to unit32? Just code: ``` package main import ( "fmt" ) func main() { uInt8 := []uint8{0,1,2,3} var uInt32 uint32 uInt32 = uint32(uInt8) fmt.Printf("%v to %v\n", uInt8, uInt32) } ``` ~>6g test.go && 6l -o test test.6 && ./test test.go:10: cannot convert uInt8 (type []uint8) to type uint32