# overriding compareTo()

Adam Johnson

Greenhorn

Posts: 3

posted 5 years ago

In Core Java book volume one, there is this tip:

i got everything in the tip except this part

What do they mean by that?

Thanks

The compareTo method of the Comparable interface returns an integer. If the objects are not equal, it does not matter what negative or positive value you return. This flexibility can be useful when comparing integer fields. For example, suppose each employee has a unique integer id, and you want to sort by employee ID number. Then you can simply return id - other.id. That value will be some negative value if the first ID number is less than the other, 0 if they are the same ID, and some positive value otherwise. However, there is one caveat: The range of the integers must be small enough that the subtraction does not overflow. If you know that the IDs are not negative or that their absolute value is at most(Integer.MAX_VALUE - 1) / 2, you are safe.

i got everything in the tip except this part

that their absolute value is at most(Integer.MAX_VALUE - 1) / 2

What do they mean by that?

Thanks

Adam Johnson

Greenhorn

Posts: 3

posted 5 years ago

Thank you Martin for your answer

But lets say we change the code to

so large becomes 2147483647 and small is -2147483647 obviously the result should be 0, but instead it yields -2

How is that possible?

I know that when we add 1 to Integer.MAX_VALUE we get -2147483648 hence it overflow the variable by 1.

but in the subtraction case above shouldn't it go back to 0 position?

But lets say we change the code to

so large becomes 2147483647 and small is -2147483647 obviously the result should be 0, but instead it yields -2

How is that possible?

I know that when we add 1 to Integer.MAX_VALUE we get -2147483648 hence it overflow the variable by 1.

but in the subtraction case above shouldn't it go back to 0 position?

Campbell Ritchie

Sheriff

Posts: 49813

69

posted 5 years ago

Welcome to the Ranch

This is a standard problem with integer arithmetic called overflow; you can overflow positively or negatively. The largest number you can fit into two's complement arithmetic in 8 bits is 0111_1111 (decimal = +127), but if you add 1 you get 1000_0000, which in two's complement is -128. Similarly if you subtract 1 from 1000_0000 (-128) you get 0111_1111 (+127).

This is a standard problem with integer arithmetic called overflow; you can overflow positively or negatively. The largest number you can fit into two's complement arithmetic in 8 bits is 0111_1111 (decimal = +127), but if you add 1 you get 1000_0000, which in two's complement is -128. Similarly if you subtract 1 from 1000_0000 (-128) you get 0111_1111 (+127).

posted 5 years ago

large + small == 0. large - small == 2147483647 - -2147483647 == 2147483647 + 2147483647 == 4294967294. Since ints can only range from -2147483648 to 2147483647 this number overflows to -2.

For comparing ints I usually use the following: In other words, -1 if value1 < value2, 1 if value1 > value2 and 0 if value1 == value2.

Adam Johnson wrote:so large becomes 2147483647 and small is -2147483647 obviously the result should be 0, but instead it yields -2

large + small == 0. large - small == 2147483647 - -2147483647 == 2147483647 + 2147483647 == 4294967294. Since ints can only range from -2147483648 to 2147483647 this number overflows to -2.

For comparing ints I usually use the following: In other words, -1 if value1 < value2, 1 if value1 > value2 and 0 if value1 == value2.

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