Uncomparable Puzzles in Java
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Join For FreeThis is a few puzzles for you to solve in Java.
b == a is true (long) b < a is true d < c is true d < (long) c is false e <= f is true e >= f is true e == f is false x == y is false x.doubleValue() == y.doubleValue() is true x.equals(y) is false x.compareTo(y) == 0 is true
See if you can work out why this code produces the output above (if you use StackOverflow, I will know ;)
long a = (1L << 54) + 1; double b = a; System.out.println("b == a is " + (b == a)); System.out.println("(long) b < a is " + ((long) b < a)); double c = 1e19; long d = 0; d += c; System.out.println("\nd < c is " + (d < c)); System.out.println("d < (long) c is " + (d < (long) c)); Double e = 0.0; Double f = 0.0; System.out.println("\ne <= f is " + (e <= f)); System.out.println("e >= f is " + (e >= f)); System.out.println("e == f is " + (e == f)); BigDecimal x = new BigDecimal("0.0"); BigDecimal y = BigDecimal.ZERO; System.out.println("\nx == y is " + (x == y)); System.out.println("x.doubleValue() == y.doubleValue() is " + (x.doubleValue() == y.doubleValue())); System.out.println("x.equals(y) is " + x.equals(y)); System.out.println("x.compareTo(y) == 0 is " + (x.compareTo(y) == 0));
A solution will be posted this week.
Bonus puzzle: The Shrinking collections.
List<BigDecimal> bds = Arrays.asList(new BigDecimal("1"), new BigDecimal("1.0"), new BigDecimal("1.00"), BigDecimal.ONE); System.out.println("bds.size()= " + bds.size()); Set<BigDecimal>bdSet = new HashSet<>(bds); System.out.println("bdSet.size()= " + bdSet.size()); Set<BigDecimal>bdSet2 = new TreeSet<>(bds); System.out.println("bdSet2.size()= " + bdSet2.size());
prints
bds.size()= 4 bdSet.size()= 3 bdSet2.size()= 1
Java (programming language)
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