“double, Double, HashSet” in Java

Question

Please see the code below.

"Double"(upper D) is used in HashSet, and "double" is used for x1, x2 and x3. After x1 is added to the HashSet, x2 cannot be added, but x3 can! Why??????

Thanks in advance :)

HashSet<Double> hs = new HashSet<Double>();
double x1, x2, x3;

x1 = (double)0/1;
System.out.println(hs.add(x1)); //true

x2 = (double)0/2;
System.out.println(hs.add(x2)); //false

x3 = (double)0/-1;
System.out.println(hs.add(x3)); //true

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And if you add "0.0 +" for x1, x2 and x3, the result is as follows.

x1 = 0.0 + (double)0/1;
System.out.println(hs.add(x1)); //true

x2 = 0.0 + (double)0/2;
System.out.println(hs.add(x2)); //false

x3 = 0.0 + (double)0/-1;
System.out.println(hs.add(x3)); //false

Answer 1

Try this to understand difference:

HashSet<Double> hs = new HashSet<Double>();
double x1, x2, x3;

x1 = (double)0/1;
System.out.println(x1 + " "+ hs.add(x1)); //true

x2 = (double)0/2;
System.out.println(x2 + " " + hs.add(x2)); //false

x3 = (double)0/-1;
System.out.println(x3 + " " + hs.add(x3)); //true

Basically doubles are signed and 0/-1 will be evaluated as -0.0 instead of 0.0 by x1 or x2.

Answer 2

HashSet<E> only allows unique values, unlike List<E> .

x1 = (double)0/1; equals to 0.0 , same as x2 = 0.0 + (double)0/2; . While x3 = (double)0/-1; equals to -0.0 . This is why you can add the first and third element but not the second.

When to use HashSet then? What is the pro of it over List ?

That has already been answered here HashSet vs. List performance , which I recommend you to read.

Some minor things of your code.

• I see no reason to mix Double and double, stick to one for easier reading. Java : different double and Double in comparison

• Normally one would write it as Set<Double> hs = new HashSet<>(); Note that it is only supported to use the Diamond Operator in Java 7+. You can read more of it here What is the point of the diamond operator in Java 7? .