Java using Integer.parseInt for comparison [duplicate]

Question

Integer x1 = Integer.parseInt("4");
Integer y1 = Integer.parseInt("4");
Integer x2 = Integer.parseInt("444");
Integer y2 = Integer.parseInt("444");

System.out.println(x1==y1); // true
System.out.println(x2==y2); // false ???

Integer a1 = Integer.valueOf("4");
Integer b1 = Integer.valueOf("4");
Integer a2 = Integer.valueOf("444");
Integer b2 = Integer.valueOf("444");

System.out.println(a1==b1); // true
System.out.println(a2==b2); // false

I understand why the third and fourth output print true and false. This is because valueOf returns an object, and wrapper classes cache objects that have values in the range of -128 to 127. If valueOf is passed any value in that range, it should reuse the object in the cache. Otherwise, it will create a new object.

Now, why does the second output print out false? I thought parseInt returns a primitive, not an object like valueOf does.

@TimBiegeleisen. I disagree this is a duplicate of that one. It does explain you should compare the .intValue() of the Integer-objects, and also gives additional information about it in general. But it fails to answer the actual question asked by the OP. I've therefore voted to re-open. — Kevin Cruijssen
– Kevin Cruijssen, Commented Jul 31, 2018 at 11:06
@KevinCruijssen Refer to the added link discussing the Integer constant pool, which explains the above question. This gets asked over and over again. — Tim Biegeleisen
– Tim Biegeleisen, Commented Jul 31, 2018 at 11:08
As to answer the question. In the source code of Integer the .valueOf method does the following: return new Integer(parseInt(s, 10, false));. As you can see, it uses .parseInt internally, and putting it in an Integer-constructor to create an Integer-instance. .parseInt does indeed return a primitive: return parseInt(s, 10, false);. However, since you save it in an Integer field, autoboxing occurs, which implicitly behaves somewhat something similar to the use of the constructor in the .valueOf method. — Kevin Cruijssen
– Kevin Cruijssen, Commented Jul 31, 2018 at 11:11

Eran · Accepted Answer · 2018-07-31 10:58:47Z

3

I thought parseInt returns a primitive, not an object like valueOf does.

parseInt returns an int but you assign it to an Integer variable, which causes auto-boxing. Since an int whose value is 444 is being auto-boxed two times, each time a new Integer instance is created (since the Integer cache cannot be used for that value), so comparing them with == returns false.

If you'll assign the output of parseInt to an int, the comparisons will return true in both cases:

int x1 = Integer.parseInt("4");
int y1 = Integer.parseInt("4");
int x2 = Integer.parseInt("444");
int y2 = Integer.parseInt("444");

System.out.println(x1==y1); // true
System.out.println(x2==y2); // true

answered Jul 31, 2018 at 10:58

Eran

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