How do I compare two Integers

Question

I have to compare two Integer objects  not int   What is the canonical way to compare them   Integer x       Integer y         I can think of this   if  x    y     The    operator only compares references  so this will only work for lower integer values  But perhaps auto-boxing kicks in      if  x equals y      This looks like an expensive operation  Are there any hash codes calculated this way   if  x intValue      y intValue      A little bit verbose     EDIT  Thank you for your responses  Although I know what to do now  the facts are distributed on all of the existing answers  even the deleted ones     and I don t really know  which one to accept  So I ll accept the best answer  which refers to all three comparison possibilities  or at least the first two

User · Answer

I would go with x equals y  because that s consistent way to check equality for all classes   As far as performance goes  equals is actually more expensive because it ends up calling intValue     EDIT  You should avoid autoboxing in most cases  It can get really confusing  especially the author doesn t know what he was doing  You can try this code and you will be surprised by the result   Integer a   128  Integer b   128   System out println a  b

User · Answer

Use the equals method  Why are you so worried that it s expensive

User · Answer

This is what the equals method does   public boolean equals Object obj        if  obj instanceof Integer            return value      Integer obj  intValue              return false      As you can see  there s no hash code calculation  but there are a few other operations taking place there   Although x intValue      y intValue   might be slightly faster  you re getting into micro-optimization territory there   Plus the compiler might optimize the equals   call anyway  though I don t know that for certain   I generally would use the primitive int  but if I had to use Integer  I would stick with equals

User · Answer

Compare integer and print its value in value ascending or descending order  All you have to do is implements Comparator interface and override its compare method and compare its value as below    Override public int compare Integer o1  Integer o2        if  ascending            return o1 intValue   - o2 intValue          else           return o2 intValue   - o1 intValue

User · Answer

I just encountered this in my code and it took me a while to figure it out  I was doing an intersection of two sorted lists and was only getting small numbers in my output  I could get it to work by using  x - y    0  instead of  x    y  during comparison

User · Answer

Minor note  since Java 1 7 the Integer class has a static compare Integer  Integer  method  so you can just call Integer compare x  y  and be done with it  questions about optimization aside    Of course that code is incompatible with versions of Java before 1 7  so I would recommend using x compareTo y  instead  which is compatible back to 1 2

User · Answer

The Integer class implements Comparable lt Integer gt   so you could try   x compareTo y     0   also  if rather than equality  you are looking to compare these integers  then   x compareTo y   lt  0 will tell you if x is less than y   x compareTo y   gt  0 will tell you if x is greater than y   Of course  it would be wise  in these examples  to ensure that x is non-null before making these calls

User · Answer

if  x equals y         This looks like an expensive operation  Are there any hash codes calculated this way    It is not an expensive operation and no hash codes are calculated  Java does not magically calculate hash codes  equals      is just a method call  not different from any other method call   The JVM will most likely even optimize the method call away  inlining the comparison that takes place inside the method   so this call is not much more expensive than using    on two primitive int values   Note  Don t prematurely apply micro-optimizations  your assumptions like  this must be slow  are most likely wrong or don t matter  because the code isn t a performance bottleneck

User · Answer

equals  is it  To be on the safe side  you should test for null-ness   x    y     x    null  amp  amp  x equals y     the x  y tests for null  null  which IMHO should be true   The code will be inlined by the JIT if it is called often enough  so performance considerations should not matter   Of course  avoiding  Integer  in favor of plain  int  is the best way  if you can    Added   Also  the null-check is needed to guarantee that the equality test is symmetric -- x equals y  should by the same as y equals x   but isn t if one of them is null

[java] How do I compare two Integers?

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