Decoding data from a byte slice to Uint32 - go

Question

package main
import (
"bytes"
"encoding/binary"
"fmt"
)
func main() {
aa := uint(0xFFFFFFFF)
fmt.Println(aa)
byteNewbuf := []byte{0xFF, 0xFF, 0xFF, 0xFF}
buf := bytes.NewBuffer(byteNewbuf)
tt, _ := binary.ReadUvarint(buf)
fmt.Println(tt)
}
Need to convert 4 bytes array to uint32 but why the results are not same ?
go verion : beta 1.1

Answer 1

You can do this with one of the ByteOrder objects from the encoding/binary package. For instance:
package main
import (
"encoding/binary"
"fmt"
)
func main() {
aa := uint(0x7FFFFFFF)
fmt.Println(aa)
slice := []byte{0xFF, 0xFF, 0xFF, 0x7F}
tt := binary.LittleEndian.Uint32(slice)
fmt.Println(tt)
}
If your data is in big endian format, you can instead use the same methods on binary.BigEndian.

Answer 2

tt := uint32(buf[0])<<24 | uint32(buf[1])<<16 | uint32(buf[2]) <<8 |
uint32(buf[3])
for BE or
tt := uint32(buf[0]) | uint32(buf[1])<<8 | uint32(buf[2]) <<16 |
uint32(buf[3]) <<24
for LE.
[u]varint is a different kind of encoding (32 bit numbers can have as much as 5 bytes in the encoded form, 64 bit numbers up to 10).
No need to create a buffer for []byte. Use Varint or Uvarint directly on the byte slice instead.
You're throwing away the error returned by the function. The second result indicates how many bytes were read or if there was a problem. There is a problem while decoding 0xff, 0xff, 0xff, 0xff as an uvarint.

Answer 3

Here is how to use the encoding/binary package to do what you want. Note that you don't want to use any of the var functions as those do variable length encoding.
Playground version
package main
import (
"bytes"
"encoding/binary"
"fmt"
"log"
)
func main() {
aa := uint(0xFFFFFF0F)
fmt.Println(aa)
tt := uint32(0)
byteNewbuf := []byte{0x0F, 0xFF, 0xFF, 0xFF}
buf := bytes.NewBuffer(byteNewbuf)
err := binary.Read(buf, binary.LittleEndian, &tt)
if err != nil {
log.Fatalf("Decode failed: %s", err)
}
fmt.Println(tt)
}
Result is
4294967055
4294967055

Answer 4

Numeric types
byte alias for uint8
Since byte is an alias for uint8, your question, "Need to convert 4 bytes array to uint32", has already been answered:
How to convert [4]uint8 into uint32 in Go?
Package binary
[Uvarints and] Varints are a method of encoding integers using one
or more bytes; numbers with smaller absolute value take a smaller
number of bytes. For a specification, see
http://code.google.com/apis/protocolbuffers/docs/encoding.html.
Since Uvarints are a peculiar form of integer representation and storage, you should only use the ReadUvarint function on values that have been written with the Uvarint function.
For example,
package main
import (
"bytes"
"encoding/binary"
"fmt"
)
func main() {
buf := make([]byte, 10)
x := uint64(0xFFFFFFFF)
fmt.Printf("%2d %2d %v\n", x, len(buf), buf)
n := binary.PutUvarint(buf, x)
buf = buf[:n]
fmt.Printf("%2d %2d %v\n", x, len(buf), buf)
y, err := binary.ReadUvarint(bytes.NewBuffer(buf))
if err != nil {
fmt.Println(err)
return
}
fmt.Printf("%2d %2d %v\n", y, len(buf), buf)
}
Output:
4294967295 10 [0 0 0 0 0 0 0 0 0 0]
4294967295 5 [255 255 255 255 15]
4294967295 5 [255 255 255 255 15]