Ruby logical operator

Question

I don't know what makes a difference here.

a = 24
b = 60
comp1 = a > 42 or b == 60
comp1 # => false
comp2 = (a > 42 or b == 60)
comp2 # => true

Could someone explain what's going on and why the return values are different?

You may be [confusing `or` with `||` here.](http://stackoverflow.com/questions/2083112/difference-between-or-and-in-ruby) — Michael Berkowski, Jun 25 '15 at 17:32
`or` and `and` are meant to be used as control flow operators. Whereas `||` and `&&` are meant to be logical operators. — engineersmnky, Jun 25 '15 at 17:35
@engineersmnky Given that every expression in Ruby has an evaluated value, that does not make any difference. — sawa, Jun 25 '15 at 17:36
@sawa: semantics, man. Conveying message and all that. Then again, lowered precedence helps with using for control flow. — Sergio Tulentsev, Jun 25 '15 at 17:37
Helps to remember that `and, or` are sometimes referred to as logical _composition_ operators. — Michael Berkowski, Jun 25 '15 at 17:41
@SergioTulentsev Control flow operators do mean something (and more than semantically) even if @sawa doesn't think so. Take this for example `a = 12 and a += 1 #=> 13` vs `b =12 && b += 1 #=> NoMethodError: undefined method '+' for nil:NilClass` this is because the first version says assign `a` and if `a` is not falsey then evaluate `a += 1` the second version the assingment will occur after the logical expression is evaluated and since `12` is truthy it then moves on to `b += 1` but `b` is still `nil` so it raises an error. — engineersmnky, Jun 25 '15 at 18:00

score 4 · Accepted Answer · answered Jun 25 '15 at 17:31

This is due to the strength of the operator binding, as operators are applied in a very particular order.

or is very loose, it has the lowest priority. The || operator is very strong, the opposite of that. Note how in that table || comes before =, but or comes after? That has implications.

From your example:

comp1 = a > 42 or b == 60

This is how Ruby interprets this:

(comp1 = (a > 42)) or (b == 60)

As such, the entire statement returns true but comp1 is assigned false because it doesn't capture the whole thing.

So to fix that, just use the strong binding version:

comp1 = a > 42 || b == 60
# => true

Sergio Tulentsev · Answer 2 · 2015-06-25T17:39:03.127

3

It has all to do with operator precedence. or has lower priority than =, so

comp1 = a > 42 or b == 60

is executed as

(comp1 = a > 42) or (b == 60)

You need to enforce precedence by parentheses. Or be a good ruby coder and never* use and/or (use &&/|| instead)

_{* never, unless you know what you're doing. A rule of thumb is: &&/|| for logical operations, and/or - for control flow.}

edited Jun 25 '15 at 17:39

answered Jun 25 '15 at 17:31

Sergio Tulentsev

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Thank you for adding the `*` to *never* and for explaining that `and` and `or` are for control flow not logical operations. – engineersmnky Jun 25 '15 at 17:36

score 1 · Answer 3 · answered Jun 25 '15 at 17:32

In Ruby, assignment (=), has higher precedence than the written or operator, so the first line is interpreted as this:

(comp1 = a > 42) or (b == 60)

That means that comp1 is being assigned the value of a > 42, which is obviously false. The parenthesis in the second expression resolve the issue.

In general, in Ruby, you use || instead of or, and likewise, && in place of and.

Ruby logical operator

3 Answers3