Java - fun converting a string (4 chars) to int and back

问题

Please don't ask why but I have to store a string (max 4 char) in an integer value (so 4 bytes).

First I wrote this and it works:

String value = "AAA";
int sum = IntStream.range(0, value.length())
                   .limit(4)
                   .map(i -> value.charAt(i) << (i * 8))
                   .sum();

System.out.println(sum);

I was unable to think a functional solution for the way back.

StringBuffer out = new StringBuffer();
while (sum > 0) {
    int ch = sum & 0xff;
    sum >>= 8;
    out.append((char) ch);
}

Any idea to write the way back (to "AAA") in a functional way?

回答1:

That's pretty similar to yours, but maybe will be helpful

    String str = IntStream.iterate(sum, s -> s > 0, s -> s >> 8)
            .mapToObj(s -> String.valueOf((char)(s & 0xff)))
            .collect(Collectors.joining());

回答2:

It may not be the best way to convert it back to a String using streams; your existing while loop seems cleaner and may be more performant. However, there is a streams solution.

String result = IntStream.range(0, 4)
        .map(i -> (value >>> 24 - (i * 8)) & 0xFF)
        .dropWhile(i -> i == 0)
        .mapToObj(v -> String.valueOf( (char) v))
        .collect(Collectors.joining());

This generates some byte indices, takes the specific byte value, maps it to a char, them joins them back to a String. It also drops leading 0 values using the Java 9+ method dropWhile.

I've corrected this to yield the proper order by adding the "24 -" to grab the most significant byte first.

With a value of 0x10101 * 65, this returns "AAA". With a value of 0x10101 * 65 + 1, this returns "AAB".

回答3:

One solution could be:

StringBuilder out =
    IntStream.iterate(sum, s -> s > 0, s -> s >> 8)
             .mapToObj(s -> (char) (s & 0xff))
             .collect(StringBuilder::new, StringBuilder::append, StringBuilder::append);

System.out.println(out.toString());

Note: I used a three-argument IntStream.iterate overload that is available starting from Java 9.

回答4:

it's basically a char[4] that's being transferred to an int then back. Realistically I would imagine this would work

public static String getStr(int encoded){
    return ((char)(encoded >> 24) + ((char)((encoded >> 16) 0xFF)
            + ((char)((encoded >> 8) & 0xFF) + ((char)((encoded) & 0xFF)
}

回答5:

Thanks to all for your contribution. I've found another solution to share with you:

String str = IntStream.iterate(sum, s -> s >> 8)
                      .limit(4)
                      .filter(s -> s > 0)
                      .mapToObj(s -> String.valueOf((char) (s & 0xff)))
                      .collect(Collectors.joining());

回答6:

Created a string of shift values to go thru the sum.
Mask off each character.
Convert to a string.
And join with a collector.

   String str = IntStream.of(0, 8, 16, 24)
            .mapToObj(i->(char)(sum>>i&0xff)+"")
             .collect(Collectors.joining());

来源：https://stackoverflow.com/questions/56657840/java-fun-converting-a-string-4-chars-to-int-and-back

标签

java

functional-programming