Efficient Go serialization of struct to disk

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情书的邮戳 2020-12-01 15:06

I\'ve been tasked to replace C++ code to Go and I\'m quite new to the Go APIs. I am using gob for encoding hundreds of key/value entries to disk pages but the gob encoding h

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  • 2020-12-01 15:33

    If you zip a file named a.txt containing the text "hello" (which is 5 characters), the result zip will be around 115 bytes. Does this mean the zip format is not efficient to compress text files? Certainly not. There is an overhead. If the file contains "hello" a hundred times (500 bytes), zipping it will result in a file being 120 bytes! 1x"hello" => 115 bytes, 100x"hello" => 120 bytes! We added 495 byes, and yet the compressed size only increased by 5 bytes.

    Something similar is happening with the encoding/gob package:

    The implementation compiles a custom codec for each data type in the stream and is most efficient when a single Encoder is used to transmit a stream of values, amortizing the cost of compilation.

    When you "first" serialize a value of a type, the definition of the type also has to be included / transmitted, so the decoder can properly interpret and decode the stream:

    A stream of gobs is self-describing. Each data item in the stream is preceded by a specification of its type, expressed in terms of a small set of predefined types.

    Let's return to your example:

    var buf bytes.Buffer
    enc := gob.NewEncoder(&buf)
    e := Entry{"k1", "v1"}
    enc.Encode(e)
    fmt.Println(buf.Len())
    

    It prints:

    48
    

    Now let's encode a few more of the same type:

    enc.Encode(e)
    fmt.Println(buf.Len())
    enc.Encode(e)
    fmt.Println(buf.Len())
    

    Now the output is:

    60
    72
    

    Try it on the Go Playground.

    Analyzing the results:

    Additional values of the same Entry type only cost 12 bytes, while the first is 48 bytes because the type definition is also included (which is ~26 bytes), but that is a one-time overhead.

    So basically you transmit 2 strings: "k1" and "v1" which are 4 bytes, and the length of strings also has to be included, using 4 bytes (size of int on 32-bit architectures) gives you the 12 bytes, which is the "minimum". (Yes, you could use a smaller type for length, but that would have its limitations. A variable-length encoding would be a better choice for small numbers, see encoding/binary package.)

    All in all, encoding/gob does a pretty good job for your needs. Don't get fooled by initial impressions.

    If this 12 bytes for one Entry is too "much" for you, you can always wrap the stream into a compress/flate or compress/gzip writer to further reduce the size (in exchange for slower encoding/decoding and slightly higher memory requirement for the process).

    Demonstration:

    Let's test the 3 solutions:

    • Using a "naked" output (no compression)
    • Using compress/flate to compress the output of encoding/gob
    • Using compress/gzip to compress the output of encoding/gob

    We will write a thousand entries, changing keys and values of each, being "k000", "v000", "k001", "v001" etc. This means the uncompressed size of an Entry is 4 byte + 4 byte + 4 byte + 4 byte = 16 bytes (2x 4 bytes text, 2x4 byte lengths).

    The code looks like this:

    names := []string{"Naked", "flate", "gzip"}
    for _, name := range names {
        buf := &bytes.Buffer{}
    
        var out io.Writer
        switch name {
        case "Naked":
            out = buf
        case "flate":
            out, _ = flate.NewWriter(buf, flate.DefaultCompression)
        case "gzip":
            out = gzip.NewWriter(buf)
        }
    
        enc := gob.NewEncoder(out)
        e := Entry{}
        for i := 0; i < 1000; i++ {
            e.Key = fmt.Sprintf("k%3d", i)
            e.Val = fmt.Sprintf("v%3d", i)
            enc.Encode(e)
        }
    
        if c, ok := out.(io.Closer); ok {
            c.Close()
        }
        fmt.Printf("[%5s] Length: %5d, average: %5.2f / Entry\n",
            name, buf.Len(), float64(buf.Len())/1000)
    }
    

    Output:

    [Naked] Length: 16036, average: 16.04 / Entry
    [flate] Length:  4123, average:  4.12 / Entry
    [ gzip] Length:  4141, average:  4.14 / Entry
    

    Try it on the Go Playground.

    As you can see: the "naked" output is 16.04 bytes/Entry, just little over the calculated size (due to the one-time tiny overhead discussed above).

    When you use flate or gzip to compress the output, you can reduce the output size to about 4.13 bytes/Entry, which is about ~26% of the theoretical size, I'm sure that satisfies you. (Note that with "real-life" data the compression ratio would probably be a lot higher as the keys and values I used in the test are very similar and thus really well compressible; still ratio should be around 50% with real-life data).

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  • 2020-12-01 15:43

    "Manual coding", you're so afraid of, is trivially done in Go using the standard encoding/binary package.

    You appear to store string length values as 32-bit integers in big-endian format, so you can just go on and do just that in Go:

    package main
    
    import (
        "bytes"
        "encoding/binary"
        "fmt"
        "io"
    )
    
    func encode(w io.Writer, s string) (n int, err error) {
        var hdr [4]byte
        binary.BigEndian.PutUint32(hdr[:], uint32(len(s)))
        n, err = w.Write(hdr[:])
        if err != nil {
            return
        }
        n2, err := io.WriteString(w, s)
        n += n2
        return
    }
    
    func main() {
        var buf bytes.Buffer
    
        for _, s := range []string{
            "ab",
            "cd",
            "de",
        } {
            _, err := encode(&buf, s)
            if err != nil {
                panic(err)
            }
        }
        fmt.Printf("%v\n", buf.Bytes())
    }
    

    Playground link.

    Note that in this example I'm writing to a byte buffer, but that's for demonstration purposes only—since encode() writes to an io.Writer, you can pass it an opened file, a network socket and anything else implementing that interface.

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  • 2020-12-01 15:47

    Use protobuf to efficiently encode your data.

    https://github.com/golang/protobuf

    Your main would look like this:

    package main
    
    import (
        "fmt"
        "log"
    
        "github.com/golang/protobuf/proto"
    )
    
    func main() {
        e := &Entry{
            Key: proto.String("k1"),
            Val: proto.String("v1"),
        }
        data, err := proto.Marshal(e)
        if err != nil {
            log.Fatal("marshaling error: ", err)
        }
        fmt.Println(data)
    }
    

    You create a file, example.proto like this:

    package main;
    
    message Entry {
        required string Key = 1;
        required string Val = 2;
    }
    

    You generate the go code from the proto file by running:

    $ protoc --go_out=. *.proto
    

    You can examine the generated file, if you wish.

    You can run and see the results output:

    $ go run *.go
    [10 2 107 49 18 2 118 49]
    
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