new SimpleDateFormat always returns same reference for a given dateFormat

Question

I was trying to replicate a bug by using the same instance of SimpleDateFormat across multiple threads. However I got stuck with another problem and did not find any answers to it.

This simple code block replicates the issues I am seeing.

DateFormat d1 = new SimpleDateFormat("ddMMyyyy");
DateFormat d2 = new SimpleDateFormat("ddMMyyyy");
DateFormat d3 = new SimpleDateFormat("ddMMyy");
System.out.println("d1 = " + d1);
System.out.println("d2 = " + d2);
System.out.println("d3 = " + d3);

The results of this operations under java 7 (1.7_0_21) is as follows

d1 = java.text.SimpleDateFormat@c5bfbc60
d2 = java.text.SimpleDateFormat@c5bfbc60
d3 = java.text.SimpleDateFormat@b049fd40

As you can see that although I am creating new objects for d1 and d2 they end up being the same reference. d3 ends up being a new instance as pattern is different.

Does java compile/runtime do this optimization? Any pointers will be helpful

Answer 1

SimpleDateFormat nor DateFormat (SimpleDateFormat superclass) nor Format (DateFormat superclass) have a toString() implemented, so the toString() from the Object class is actually executed, whose code is :

public String toString() {
    return getClass().getName() + "@" + Integer.toHexString(hashCode());
}

Now, SimpleDateFormat hashCode is generated:

public int hashCode()
{
    return pattern.hashCode();
    // just enough fields for a reasonable distribution
}

Which means that if you create numerous SimpleDateFormat instances with the same pattern, like in your case, they will have the same hashCode and hence toString() will return the same for these instances.

Moreover, as it has been spotted by rixmath, SimpleDateFormat instances with the same pattern will also be equal.