how does the groovyc work?

Question

It seems the groovy also support the compiling mode, using groovyc , If I run the following code with dynamic script calling way, I will get "String" method called.

Why I still got "String" even if I compiled the code using groovyc? The type of variable a is "Object", so I expected after compiling the code, I will get the "Object" function called.

Object a = "123"

def foo(Object a) {
    println "Object"
}

def foo(String a) {
    println "String"
}

foo(a)

Object a = "123" is like Object a=(GString) "123"... it will get dispatched to most matching method call.. — Jayan, Dec 05 '15 at 14:16
related reading: http://stackoverflow.com/questions/1572322/overloaded-method-selection-based-on-the-parameters-real-type (foo is a method on a Class that represents your script ) — Jayan, Dec 05 '15 at 14:18
Why would you expect the semantics to change after compilation? — Oliver Charlesworth, Dec 05 '15 at 15:29
@jayan: that link is about the *opposite* behaviour to what the OP is observing... — Oliver Charlesworth, Dec 05 '15 at 15:31
@Oliver Charlesworth : I posted the link hastily. Groovy has section about the diff with Java. This is one, and exact opposite behavior. — Jayan, Dec 05 '15 at 15:42

score 2 · Accepted Answer · edited May 23 '17 at 12:30

Here is relevant section from groovy-docs

In Groovy, the methods which will be invoked are chosen at runtime. This is called runtime dispatch or multi-methods. It means that the method will be chosen based on the types of the arguments at runtime. In Java, this is the opposite: methods are chosen at compile time, based on the declared types.

There is a sample code in '2. Multi-methods' section, not copying here.

Finally mandatory link to MrHaki's groovy goodness page on this topic

how does the groovyc work?

1 Answers1