golang：将 IPv4 和 IPv6 地址从文本格式转换为二进制格式

Question

looking to send an ipv4 address across in 4 bytes, and ipv6 address in 16 bytes - What's similar to inet_pton() in go?

struct sockaddr_in sa;
char str[INET_ADDRSTRLEN];
inet_pton(AF_INET, "192.0.2.33", &(sa.sin_addr));

struct sockaddr_in6 sa;
char str[INET6_ADDRSTRLEN];
inet_pton(AF_INET6, "2001:db8:8714:3a90::12", &(sa.sin6_addr));

I know of https://play.golang.org/p/jn8t7zJzT5v - that is looking complicated for IPV6 addresses though.

Thanks!

Answer 1

net.ParseIP() will take an IPv4 or IPv6 formatted string and return a net.IP containing the IP address.

The net.IP is what you'll need to feed to most other Go functions, such as to make a connection to the host.

Note that unlike most Go functions which return an error, net.ParseIP() simply returns nil if the string could not be parsed into an IP address.

https://play.golang.org/p/Cgsrgth7JKY

Answer 2

Either you can use the net package: `

a := net.ParseIP("127.0.0.1")
fmt.Printf("%b %s", net.IP.To4(a))

` https://play.golang.org/p/KzqYpk9OBh8

Or, you can split the IP and convert each value into an integer using strconv.Atoi(), and thereafter convert each integer into byte()

`

ipString := "127.0.0.1"

octets := strings.Split(ipString, ".")

octet0, _ := strconv.Atoi(octets[0])
octet1, _ := strconv.Atoi(octets[1])
octet2, _ := strconv.Atoi(octets[2])
octet3, _ := strconv.Atoi(octets[3])

b := [4]byte{byte(octet0),byte(octet1),byte(octet2),byte(octet3)}

fmt.Printf("%s has 4-byte representation of %b\n", ipString, b)

` https://play.golang.org/p/2F3bC0df9wB