Library Puzzlers - Java Puzzlers: Traps, Pitfalls, and Corner Cases - page 175

}

B. When it comes to parsing a string into tokens, many programmers' thoughts turn immediately

to StringTokenizer . This is most unfortunate, as StringTokenizer became obsolete as of
release 1.4, when regular expressions were added to the platform ( java.util.regex ). If you
tried to solve this puzzle with StringTokenizer , you quickly realized that it isn't a very good
fit. With regular expressions, it's a snap. To solve this puzzle in one line, use the convenience
method String.split , which takes a regular expression describing the token delimiter. If you
haven't used regular expressions before, they may look a bit cryptic, but they're amazingly
powerful and well worth learning:

static String[] parse(String string) {

return string.split(",\\s*");

}

C. This is a trick question. You don't even have to write a method. The method is provided for

you in release 5.0 and later releases, and is called Arrays.deepToString . If you pass it an
array of object references, it returns a nice string representation. It can deal with nested arrays
and even circular references, where an array element refers to the enclosing array, directly or
indirectly. In fact, the Arrays class in release 5.0 provides a whole family of toString ,
equals , and hashCode methods that allow you to print, compare, or hash the contents of any
array of primitives or object references.

D. In order to solve this puzzle in one line, you need to know that a whole family of bit-twiddling

methods were added to the platform in release 5.0. The wrapper classes for the integral types
( Integer , Long , Short , Byte , and Char ) now support common bit-manipulation operations,
including highestOneBit , lowestOneBit , numberOfLeadingZeros , numberOfTrailingZeros ,
bitCount , rotateLeft , rotateRight , reverse , signum , and reverseBytes . In this case, what
you need is Integer.bitCount , which returns the number of set bits in an int value:

static boolean hasMoreBitsSet(int i, int j) {

return (Integer.bitCount(i) > Integer.bitCount(j));

}