If you have return, it is an explicit return which can be at any point in a method, but the last expression of any method is always returned as an implicit return, so no return is ever needed for the last expression in a Ruby method.

There is an expression "ab" -> add (str (normal ('a')), normal ('b')) in the Some examples section but this expession gives "(ab)" instead of "ab".

Could someone explain this above?

Why the following code in c++ produced the following output? Is this correct?

code:

std::cout << pretty( add(any(), normal('a'))) << std::endl;

std::cout << pretty( add(normal('a'), any())) << std::endl;

std::cout << pretty( add(normal('a'), normal('b'))) << std::endl;

std::cout << pretty( add(normal('b'), normal('a'))) << std::endl;

output:

.

a

a

b

Is this ok?

new Void() in Java

??? The Java equivalent is null, like in C# and Javascript. The Void type in Java has no public constructor, so you can't call new Void() and you can't return an instance of Void. It's just there to represent the void keyword as something you can interact with.

Probably not

Parsing "((cao.z)(b|q))": expected:<((cao.z)(b|q))> but was:<(cao.z(b|q))>

This two answers have same RPN, so its impossible to do with RPN

Is it even possible to do with reverse polish notation? Am stuck with the parenthesis problem (abc) != ((ab)c)

Funny, this undef_method delegates #== and #! to @obj and makes this the shortest solution!

If your paretheses are matched, it should be valid.

Kotlin:
Str(listOf(Normal('a'), Normal('b')))
Parsing "ab": expected:<[ab]> but was:<[(ab)]>

I was feeling proud of myself, but then I saw your solution, simply breathtaking.

same! xD

but isn't that the very same for any Kata in any language?