Generating byte array within Go

Question

I am attempting to port some Python code over to Go.

In my Python code I do:

print(b"\x02" + (3).to_bytes(4, "big", signed=True))

This returns me []byte{0x2, 0x0, 0x0, 0x0, 0x3}

This adds 2 to the start of my byte array and then 3 on the end with a prefix of 3 0's.

In GO I'm doing

digest = append(digest, byte(0x02))
    err := binary.Write(buf, binary.BigEndian, int64(3))
    if err != nil {
        return nil, err
    }
digest = append(digest, buf.Bytes()...)
return digest

This returns []byte{0x2, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x3}

In Python I can choose the length which seems to be the issue, is there something similar in Go when converting an int to bytes?

Thanks!

8 byte is used for int64, you can use int32 for 4 byte. You will find details here golang.org/pkg/builtin — Eklavya
– Eklavya, Commented May 18, 2020 at 16:11
a 64 bit value is 8 bytes. If you only want 4 bytes, use a 32bit value. — Mr_Pink
– Mr_Pink, Commented May 18, 2020 at 16:11
Thanks both! If you'd like to add this an answer I'll accept. — N P
– N P, Commented May 20, 2020 at 10:05

schilli · Accepted Answer · 2021-05-27 21:30:03Z

1

Like suggested in some comments, too, you can just use a different data type e.g. int32 instead of int64.

digest = append(digest, byte(0x02))
    err := binary.Write(buf, binary.BigEndian, int32(3))
    if err != nil {
        return nil, err
    }
digest = append(digest, buf.Bytes()...)
return digest

This will produce the desired outcome []byte{0x2, 0x0, 0x0, 0x0, 0x3}.

answered May 27, 2021 at 21:30

schilli

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