Does the || operator evaluate the second argument even if the first argument is true?

Question

I'm trying to evaluate the expression (a=10) || (rr=20) while the rr variable is not defined

so typing rr in the ruby console before evaluating the previous expression returns

rr
NameError: undefined local variable or method `rr' for main:Object
from (irb):1
from :0

When I write the expression (a=10) || (rr=20) it returns 10, and when I write rr afterwards it says nil

(a=10) || (rr=20)
rr  # => nil

so, why is this happening? Shouldn't rr be defined only if the second argument of the || operator is evaluated, which should be never based on the documentation?

Answer 1

This happens because the ruby interpreter defines a variable when it sees an assignment to it (but before it executes the actual line of code). You can read more about it in this answer.

Boolean OR (||) expression will evaluate to the value of left hand expression if it is not nil and not false, else || will evaluate to the value of right hand expression.

In your example the ruby interpreter sees an assignment to a and rr (but it doesn't execute this line yet), and initializes (defines, creates) a and rr with nil. Then it executes the || expression. In this || expression, a is assigned to 10 and 10 is returned. r=20 is not evaluated, and rr is not changed (it is still nil). This is why in the next line rr is nil.

Answer 2

As @DOC said, && and || are known as short circuited conditional operators.

In case of ||, if the left part of || expression returns true, the right part won't be executed. That means the right part will be executed only if the left part of || expression returns false.

In case of &&, right part of the&&expression will be executed only if left part of && returns true.

In the given scenario (a=10) || (rr=20), rr=20 won't be executed since the ruby expression a=10 returns true. Note that in ruby assignment expression returns true except nil and false.