问题
I was trying to construct functional program for parsing IP address. I am seeing an error. I wanted a simpler code which differentiates ipv4 to ipv6. Here is the JAVA code.
import java.util.regex.Pattern;
class Solution {
String chunkIPv4 = "([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])";
Pattern pattenIPv4 =
Pattern.compile("^(" + chunkIPv4 + "\\.){3}" + chunkIPv4 + "$");
String chunkIPv6 = "([0-9a-fA-F]{1,4})";
Pattern pattenIPv6 =
Pattern.compile("^(" + chunkIPv6 + "\\:){7}" + chunkIPv6 + "$");
public String validIPAddress(String IP) {
if (pattenIPv4.matcher(IP).matches()) return "IPv4";
return (pattenIPv6.matcher(IP).matches()) ? "IPv6" : "Neither";
}
}
回答1:
Assuming your scala solution that you wrote in the comment has the following:
def validIPAddress(IP: String): String = {
if (pattenIPv4.matcher(IP).matches()) "IPv4"
if (pattenIPv6.matcher(IP).matches()) "IPv6"
else "Neither"
}
The first if
line will be evaluated but will not return without a return
keyword, so it will fall through the next conditional.
You can fix that in two ways, one is to add return
:
if (pattenIPv4.matcher(IP).matches()) return "IPv4"
or maybe better add an else
to the second line, so you can avoid the return
as the whole thing will be evaluated as a single expression:
def validIPAddress(IP: String): String = {
if (pattenIPv4.matcher(IP).matches()) "IPv4"
else if (pattenIPv6.matcher(IP).matches()) "IPv6"
else "Neither"
}
Also, as a side note, all those var
s can be val
s since you are not mutating them, and it's a good practice in scala to have the guarantee that they will always have the same value.
来源:https://stackoverflow.com/questions/60496856/need-a-scala-functional-code-for-validating-ipv4-and-ipv6