Expression Language lambda type inference

Question

Normally I would do:

Function<Integer, Integer> a = b -> b * 2;
System.out.println(a.apply(3)); // prints 6

I was amazed to see that following EL expression works:

${a = b -> b * 2; a(3)}

The result of above EL expression is 6. How can compiler determine the type when declaring a in snippet 2 but requires type information in snippet 1?

Even this compiles and executes fine:

${(b -> b * 2)(3)}

Answer 1

The EL evaluator / executor has no idea about the exact type of a in your EL expression. It just knows it should be a Number because of * operation

${a = b -> b * 2; a(3)}

Take a look at these results:

${a = b -> b * 2; a(3)}
6

${a = b -> b * 2; a('32')}
64

${a = b -> b * 2; a('32s')}
java.lang.NumberFormatException: For input string: "32s"

So only at runtime you will get an Exception, because Long.parse("32s") fails.

Look at this source code of ELArithmetic.java#211