java - do <= and >= relational operators work with Integer objects -

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• how compare 2 integers in java? 6 answers

i understand can't use == or != compare values of numeric objects , have use .equals() instead. after fair amount of searching haven't been able find statement whether or not can use other comparison operators, other suggestions use .compare() or .compareto() feel inefficient because require 2 comparisons: b, result of zero.

despite == , != comparing addresses of objects, other comparison operators appear compare numeric values. instance following code snippet:

integer = new integer(3000); integer b = new integer(3000); system.out.println("a <  b " + (a < b)); system.out.println("a <= b " + (a <= b)); system.out.println("a == b " + (a == b)); system.out.println("a >= b " + (a >= b)); system.out.println("a >  b " + (a > b)); 

produces

a <  b false <= b true == b false >= b true >  b false 

which appears indicate operators == compare value, not address of object. accepted usage use <= class of operators, or unsupported feature of compiler or something?

yes, note integer objects, not primitive int. usage of >, <, >= , <= operators not meant objects, primitives, when using of these, integer autoboxed int. while when using == in objects comparing references. use equals instead compare them.

still, note integer class has cache store integer references -128 127. means if this:

integer i1 = 127; integer i2 = 127; system.out.println(i1 == i2); 

will print true.