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Overload resolution issue

 
Greenhorn
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Could someone explain how java compiler work in this situation?


 
Vladimir Yatsevskiy
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P.S. i know about this rules.
 
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Hi Vladimit. Welcome to the Ranch!

When there are more than one method that can match a method call, the exact rules for determining precedence are a bit nasty. You can read part 15.12.2.5 of the Java Language Specification for the full details.

(Edit: posted before your second message)

But briefly, in the first example there's a rule that can be applied to decide that f(int, Integer) shoud be applied. It's more specific than f(long, Integer) because the first parameter is actually an int. The rules as defined don't give a way of deciding between the two methods in your second example.
 
Matthew Brown
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For all j from 1 to n, Tj <: Sj.


This bit from the rules, roughly speaking, says that a method is more specific if all its parameters are at least as specific (and at least one is more specific). It doesn't try to cope with the situation where some are more specific and some less specific.
 
Vladimir Yatsevskiy
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Matthew Brown wrote:Hi Vladimit. Welcome to the Ranch!

.



Thanks. actually my name is Vladimir. I sent request for fix this misprint. thanks for you answer.
 
Consider Paul's rocket mass heater.
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