Ruby short circuit "or"

gistfile1.txt

# First, let's explain some "normal" conditional OR behavior. 
# I'm going to use some IRB output here because IRB always prints
# an evaluation value to the screen; this will help.

# Just setting up some variables to play with
# The comment that starts with "#=>" is the result
# of the operation, and IRB prints this out for us
# automatically
1.9.3p194 :001 > a = 1
#=> 1 
1.9.3p194 :002 > b = 2
#=> 2 

# Here, you can see the expected output of 1
# That's because a was evaluated first
# But notice that IRB printed a 2, not nil (return value of puts method).

# That's becaue b, which equals 2, was the last
# thing evaluated, and therefore is what IRB prints out to us.
# It is proof that b was in fact evaluated
1.9.3p194 :003 > puts a or b
1
#=> 2 

# Now, using short circuit evaluation instead. 
# We can see that we still get the logical result from a

# But b is never even evaluated, thus IRB
# simply printed the value of evaluating, essentially
# the statement "puts a", which produces the default
# return value from puts, which is nil.
1.9.3p194 :004 > puts a || b
1
#=> nil