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function call operator? [message #1751227]

Thu, 05 January 2017 20:23

Eric Salemi

Messages: 35
Registered: September 2016
Location: Belgium

Member

Today I scoured the XTend source code looking for an operator that would do something as simple as taking an input argument, binds it to the scope of a lambda expression and returns the output of the lambda. Maybe I searched wrong but I could not find anything like that.

In the end I implemented it as a method:

def <T, R> R call(T input, Function1<T, R> lambda) {
        lambda.apply(input)
    }

And it can be used like this:

1.call[i|i+1].call[j|j-1].assertEquals(1)

I just thought it could be useful to have an operator for this.

Questions:

1) Can I use custom operators in XTend?
2) Could we add this operation to the ObjectExtensions class?

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Re: function call operator? [message #1751229 is a reply to message #1751227]

Thu, 05 January 2017 21:00

Christian Dietrich

Messages: 14665
Registered: July 2009

Senior Member

(1) no you can not introduce custom operators, you can use/overload the existing ones only (see org.eclipse.xtext.xbase.scoping.featurecalls.OperatorMapping class)
in your case you could use e.g.

class Demo {
	def x() {
		1>>>[i|i+1]
	}
	
	def <T, R> R operator_tripleGreaterThan(T input, Function1<T, R> lambda) {
        lambda.apply(input)
    }
}

(2) feel free to file a ticket at https://github.com/eclipse/xtext-extras and/or https://github.com/eclipse/xtext-lib. the base problem is to find a "free" operator for that

Twitter : @chrdietrich
Blog : https://www.dietrich-it.de

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Re: function call operator? [message #1751232 is a reply to message #1751229]

Thu, 05 January 2017 22:43

Eric Salemi

Messages: 35
Registered: September 2016
Location: Belgium

Member

When using an operator I apparently have to enclose the whole expression in parenthesis if I want to apply another operator.

For example this does not work:

def void test() {
    1 >>> [return 10] + 1
}

But this does:

def void test() {
    (1 >>> [return 10]) + 1
}

When the syntax used is a method call, the next operator can be written right after. I'm not sure I understand the reason why. Could it be related to operator precedence?

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Re: function call operator? [message #1751248 is a reply to message #1751232]

Fri, 06 January 2017 09:20

Jan Koehnlein

Messages: 760
Registered: July 2009
Location: Hamburg

Senior Member

Yes, operator precedence is fixed.
As we're using Antlr,the precedence is defined by the call hierarchy in the Xtend grammar. You cannot change that on library level.

---
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