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how is this possible please explain

 
Shashank Agarwalg
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Android Java Redhat
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Integer i1=1000;

Integer i2=1000;

if(i1!=i2) System.out.println("different objects");

if(i1.equals(i2)) System.out.println("meaningfully equal");


output

different objects //how is this possible please explain

meaningfully equal









b]Integer i3=10;

Integer i4=10;

if(i3==i4) System.out.println("same objects");

if(i3.equals(i4)) System.out.println("meaningfully equal");


output

same objects //how is this possible please explain

meaningfully equal


how if(i3==i4) and if(i1!=i2) both can be true
 
marc weber
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This is due to autoboxing. In your first example, the value being boxed is outside the range -128 to 127. In your second example, the value is within that range. As explained under JLS - 5.1.7 Boxing Conversion...
If the value p being boxed is true, false, a byte, a char in the range \u0000 to \u007f, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.
 
Shashank Agarwalg
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Thanks for the solution.
 
Jesper de Jong
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