compare

hi expert..
say:
int i = 500;
int j = 600;

String s1 = "Sad";
String s1 = "Bad";

if( i == j ) or if( s1.equals( s2) ) more faster???

please explan why?
LVL 8
Dishan FernandoSoftware Engineer / DBAAsked:
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mmuruganandamConnect With a Mentor Commented:
I am strongly saying that primitive data types are much faster than anything.  
Object are always slower than the primitive data type operations.


Here is the proof...

      public static void main(String[] args)
      {
            String a = "hello";
            String b = "whatis";
            
            int aa = 500;
            int bb = 600;
            
            long start = System.currentTimeMillis();
            for (int i = 0; i < 1000000; i++)
            {
                  if (aa == bb)
                  {
                  }
            }
            System.out.println("Integer comparison... time taken: " + (System.currentTimeMillis() - start));
            
            start = System.currentTimeMillis();
            for (int i = 0; i < 1000000; i++)
            {
                  if (a.equals(b))
                  {
                  }
            }
            System.out.println("String comparison... time taken: " + (System.currentTimeMillis() - start));
      }
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CEHJConnect With a Mentor Commented:
int will be faster - as it's a leaner operation. Why are you interested?
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mmuruganandamConnect With a Mentor Commented:
primitive data type will be faster always.

i == j is a simple computation....

String is an object and there is a method invocation will cost bit more time

So always the primitive data type operations are faster than the object comparison.


Regards,
Muruga
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CEHJCommented:
Hang on - they're actually likely to be virtually identical
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searlasConnect With a Mentor Commented:
You're comparing an operator to a method invocation.  It doesn't matter that you're calling 'equals' on the string, it could be any method.

I'll go out on a limb here and say it's highly likely that ALL operators in Java are faster than method invocations.

i.e.  i == j, i << j, i + j, i * j, i / j.....

Faster than.
obj.method(param);

Method invocation implies stack manipulation, looking up method entry point etc etc...
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searlasCommented:
(before someone points it out... I'm not including str1 + obj as an operator as that implies calls to toString... therefore has implicit method invocation...)
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CEHJCommented:
I was thinking that the VM might have a way of knowing that a String taken from the constant pool is not the same as another one taken from the constant pool, but on inspection, it looks like this is not the case. The only real way to ascertain this sort of thing is by profiling it. 'Opinions' are not very useful
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mmuruganandamCommented:
>> String b = "whatis";

String b = "whati";

with same length data with different string will increase the time taken...
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mmuruganandamCommented:
that's great :-)
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CEHJCommented:
Along the lines of what i was getting at

         String a = "hello";
         String b = "hello";

will take quite a bit less time and

            String a = "haaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
            String b = "haaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";

won't be any slower
         
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