Surprising length calculation

[thomas-mangin]

// Serialize serializes the handshake to a buffer
func (h *Handshake) Serialize() []byte {
	buf := make([]byte, len(h.Pstr)+49)
...
}

Len(h.Ptr) is always >=1, the string being 19 bytes, so it will never be a problem but reading your article I would just have expected 50. 1 + 19 + 20 + 20. This allocates 19 + 49 bytes.

[veggiedefender]

It's been a while since I've thought about this project, so you may very well be correct. My understanding of where len(h.Pstr) + 49 comes from is the fact that the handshake has:

• 1 byte for pstrlen
• len(h.Pstr) bytes for the actual pstr
• 8 bytes reserved for extensions
• 20 bytes for infohash
• 20 bytes for peer id

So we always have 1 + 8 + 20 + 20 = 49 bytes, plus the variable sized pstr. So the total buffer should be len(h.Pstr) + 49 bytes long. Does that make sense?

[thomas-mangin]

Reading things late :-)