如何将int数组快速转换为字节数组？

Question

I have a process that needs to pack a large array of int16s to a protobuf every few milliseconds. Understanding the protobuf side of it isn't critical, since all I really need is a way to convert a bunch of int16s (160-16k of them) to []byte. It's a CPU-critical operation, so I don't want to do something like this:

for _, sample := range listOfIntegers {
  protobufObject.ByteStream = append(protobufObject.Bytestream, byte(sample>>8))
  protobufObject.ByteStream = append(protobufObject.Bytestream, byte(sample&0xff))
}

(If you're interested, this is the protobuf)

message ProtobufObject {
  bytes byte_stream = 1;
  ...               = 2;
  etc.
}

There has to be a faster way to supply that list of ints as a block of memory to the protobuf. I've fiddled with the cgo library to get access to memcpy, but suspect I've been destroying an underlying go data structure because I get crashes in totally unrelated sections of code.

Answer 1

A faster version of the above code is:

protobufObject.ByteStream := make([]byte, len(listOfIntegers) * 2)
for i, n := range listOfIntegers {
  j := i * 2
  protobufObject.ByteStream[j+1] = byte(n)
  protobufObject.ByteStream[j] = byte(n>>8)
}

You can avoid copying the data when running on a big-endian architecture.

Use the unsafe package to copy the []int16 header to the []byte header. Use the unsafe package again to get a pointer to the []byte header and adjust the length and capacity for the conversion.

b = *(*[]byte)(unsafe.Pointer(&listOfIntegers))
hdr := (*reflect.SliceHeader)(unsafe.Pointer(&b))
hdr.Len *= 2
hdr.Cap *= 2
protobufObject.ByteStream = b