Initializing a Double object with a primitive double value

java autoboxing
Doug Porter picture Doug Porter · Jul 20, 2010 · Viewed 22.7k times · Source

What is happening when a java.lang.Double object is initialized without using a call to the constructor but instead using a primitive? It appears to work but I'm not quite sure why. Is there some kind of implicit conversion going on with the compiler? This is using Java 5.

public class Foo {

    public static void main(String[] args) {
        Double d = 5.1;

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

    }

}

Answer

krock picture krock · Jul 20, 2010

This is called Autoboxing which is a feature that was added in Java 5. It will automatically convert between primitive types and the wrapper types such as double (the primitive) and java.lang.Double (the object wrapper). The java compiler automatically transforms the line:

Double d = 5.1;

into:

Double d = Double.valueOf(5.1);

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