i = true and false in Ruby is true?

Question

Am I fundamentally misunderstanding Ruby here? I've been writing Ruby code for about 2 years now and just today stumbled on this...

ruby-1.8.7-p249 > i = true and false
 => false 
ruby-1.8.7-p249 > i
 => true 

Could somebody explain what's going on here please? I'm sure it's to spec, but it just seems counter intuitive to me...

Answer 1

and has lower precedence than = so i = true and false is parsed as (i = true) and false.

So the value true is assigned to i and then the return value of that operation (which is true) is anded with false, which causes the whole expression to evaluate to false, even though i still has the value true.

Answer 2

The operators && and and have different precedence, and = happens to be in between.

irb(main):006:0> i = true and false
=> false
irb(main):007:0> i
=> true
irb(main):008:0> i = true && false
=> false
irb(main):009:0> i
=> false
irb(main):010:0> 

The first is read as (i = true) and false, the second as i = (true && false).

Answer 3

Your line is parsed as

i = true and false
(i = true) and false
true and false

And of course because of i = true, i will be true afterwards.

Answer 4

As I understand your code, it is interpreted as :

• Assign true to i
• Return i and false

The results seems correct.

Answer 5

As others have elucidated above, the keyword and is used when you want to put two different statements on one line. It is just a nicer way of making your code readable.

Thus,

 i = true and false 

implies

i = true; false #(a less widely used code layout in ruby)

or which is the most straightforward way:

  
 i = true
 false 

So, the output is correct. Otherwise, if you were expecting false, then use the boolean and &&.