Why does Go net.IPv4 create a 16 byte address internally - go

Question

I noticed Go creates a 16 byte internal representation of a 4 byte IPv4 Address when using:
// IPv4 returns the IP address (in 16-byte form) of the
// IPv4 address a.b.c.d.
func IPv4(a, b, c, d byte) IP {
p := make(IP, IPv6len)
copy(p, v4InV6Prefix)
p[12] = a
p[13] = b
p[14] = c
p[15] = d
return p
}
https://golang.org/src/net/ip.go
Is there a reason that IPv4 is initially created with 16 bytes? I was doing some calculations for Broadcast and Networkaddress where I accessed the internal byte[] directly and got confused that I had to call To4() to do something like
start := binary.BigEndian.Uint32([]byte(ip))
and actually get the IPv4 Address as uint32.

Answer 1

It looks like they're using a representation that allows them to easily handle both IPv4 and IPv6 addresses with the same code. They're using a single 16-byte representation for either type of IP.
The comments at the top of the code (that you linked to) explain in pretty clearly:
// IPv4 addresses are 4 bytes; IPv6 addresses are 16 bytes.
// An IPv4 address can be converted to an IPv6 address by
// adding a canonical prefix (10 zeros, 2 0xFFs).
// This library accepts either size of byte slice but always
// returns 16-byte addresses.
You can pass in either 4-byte or 16-byte representations of IPv4 addresses, but the library will internally use 16-byte ones and always return those.

Answer 2

The 16 byte address for IPV4 is known the IPv4-mapped address.
For 1.2.3.4 it looks like 0:0:0:0:0:ffff:12:34.
IPv4-mapped addresses are one way of storing an IPv4 address inside an IPv6 address.
Seems that going straight to the IPv4-mapped address was some kind of design choice made with that package, the reasoning is not clear to me though.