Question

以 uint16 与字节数组互转为例，看一下 Golang 中 encoding/binary 包中的字节序处理与实现。可见实现并不复杂，注意字节顺序即可。

func TestEndian(t *testing.T) {
    bytes := make([]byte, 2)
    binary.LittleEndian.PutUint16(bytes, 1024) // 小端写 --> 0x0004
    binary.BigEndian.PutUint16(bytes, 1024) // 大端写 --> 0x0400
    binary.LittleEndian.Uint16(bytes) // 小端读 --> 4
    binary.BigEndian.Uint16(bytes) // 大端读 --> 1024
}

func (littleEndian) PutUint16(b []byte, v uint16) {
    _ = b[1] // early bounds check to guarantee safety of writes below
    b[0] = byte(v)
    b[1] = byte(v >> 8)
}

func (bigEndian) PutUint16(b []byte, v uint16) {
    _ = b[1] // early bounds check to guarantee safety of writes below
    b[0] = byte(v >> 8)
    b[1] = byte(v)
}

func (littleEndian) Uint16(b []byte) uint16 {
    _ = b[1] // bounds check hint to compiler; see golang.org/issue/14808
    return uint16(b[0]) | uint16(b[1])<<8
}

func (bigEndian) Uint16(b []byte) uint16 {
    _ = b[1] // bounds check hint to compiler; see golang.org/issue/14808
    return uint16(b[1]) | uint16(b[0])<<8
}