Kurt Euler wrote:
Thanks Hal. Question: What do you mean by use ||? Could you
give an example? (I’m new to Ruby and programming.)
is essentially the same as or
&& is essentially the same as and
So:
if this == that || this =~ /another/
is the same as
if this == that or this =~ /another/
History lesson: ‘and’ and ‘or’ and ‘not’ are used in
languages like BASIC, Algol, and Pascal. && and || and !
are used in languages like C, C++, and Java. Ruby happens
to know both forms.
And by the way, don’t confuse the &&/|| operators with
the &/| operators. The latter are “bitwise” – they act
on individual bits in a number. For example. 5 | 6 is 7,
since you’re doing a bitwise OR of binary 101 and 110,
giving binary 111.
Whirlwind tour of truth and falsehood in Ruby:
Two rules -
- false and nil test as false
- everything else tests as true
Example:
if this # prints “no” if this is nil or false;
puts “yes” # prints “yes” if it’s 0 or “” or
else # or 237 or “foobar” or [1,2,3] or
puts “no” # any other value imaginable.
end
Also remember that these expressions are “short-circuited” –
as soon as we know the result, we stop executing. Especially
remember this if you’re calling a method with side effects
(in other words – don’t).
Example:
x = true || foo() # foo doesn’t get called – we already
# know the result is true
x = true && foo() # foo gets called – it’s necessary in
# order to get the correct answer
x = false || foo() # foo is called
x = false && foo() # foo is not called
This makes an interesting idiom possible. Suppose we
want to assign 5 to x, but ONLY if it doesn’t already
have a value.
An unassigned variable effectively has a value of nil.
So we can say
x = x || 5
to do this. Or using the shorthand that Ruby borrowed from
C, we can say:
x ||= 5
It’s worth mentioning that this is a little strange on
second thought, but makes sense on third thought.
Variables in Ruby aren’t declared. Ruby knows something is
a variable when it sees an assignment to it.
Now in the statement x = x || 5 you would take it for
granted that the right hand side is evaluated first so that
we can assign to the variable on the left hand side. But
how can x || 5 make sense when x is undefined??? How does it
even know x is a variable?
The answer is that the interpreter “sees” the assignment to
x at parse time and realizes that x is a variable. But since
it has never been assigned, it will have the value nil. So
the right hand side evaluates to nil || 5 (which is just 5).
So then x gets its first “real” value.
Confused yet??
Ask all the questions you want, we’ll take turns on them.
Cheers,
Hal
