Effectively convert little endian byte slice to int32

I have a data stream of bytes and I'd like to get a little endian encoded int32 from four bytes. Is there a better way than to do this like the following code?

package main

func read_int32(data []byte) int32 {
    return int32(uint32(data[0]) + uint32(data[1])<<8 + uint32(data[2])<<16 + uint32(data[3])<<24)
}

func main() {
    println(read_int32([]byte{0xFE,0xFF,0xFF,0xFF})) // -2
    println(read_int32([]byte{0xFF,0x00,0x00,0x00})) // 255
}

The code above seems to work fine, but perhaps there is a built-in function in Go that I've missed or there is a super cool hack that does that in one instruction?

encoding/binary package may have what you need. Check this: http://golang.org/pkg/encoding/binary/#example_Read

Your code with modified read_int32 function could be:

package main
import (
    "bytes"
    "encoding/binary"
    "fmt"
)

func read_int32(data []byte) (ret int32) {
    buf := bytes.NewBuffer(data)
    binary.Read(buf, binary.LittleEndian, &ret)
    return
}

func main() {
    fmt.Println(read_int32([]byte{0xFE, 0xFF, 0xFF, 0xFF})) // -2
    fmt.Println(read_int32([]byte{0xFF, 0x00, 0x00, 0x00})) // 255
}

Also you can interpret big endian by replacing binary.LittleEndian with binary.BigEndian

StackExchange Code Review Q#15945, answer score: 16