[masklinn]

> Thanks for one more insight, by showing me that sending message/calling a method is a binary operation

The other way around.

A "binary message" is specifically the message of binary operators. Smalltalk also has unary messages (named messages with no parameters), and keyword messages (non-binary messages with parameters).

Its precedence rules are unary > binary > keyword, and left to right.

So

    foo bar - baz / qux quux: corge

binds as

    (((foo bar) - baz) / qux) qux: quux

You can still get into sticky situations, mostly thanks to the cascading operator ";":

    a b; c

sends the message following ";" to the receiver of the message preceding ";". So here it's equivalent to

    a b.
    a c

now consider:

    foo + bar qux; quux

This is equivalent to

    foo + bar qux.
    foo quux

Because the message which precedes the ";" is actually "+", whose receiver is "foo":

    foo + (bar qux); quux