print (-1 == -1) and (myobj.nil?)
true
print (-1 == -1) && (myobj.nil?)
false
Note, myobj.nil? returns false so, should not this always be false.
views:
62answers:
2
+1
A:
Because they have different operator precedence.
The first evaluates as:
(print (-1 == -1)) and (myobj.nil?)
It prints true, because -1 == -1, then print returns nil and nil and false is nil.
The second is equivalent to:
print ((-1 == -1) && (myobj.nil?))
It prints (true && false), which is false, then print returns nil.
Matthew Flaschen
2010-05-14 06:32:42
but why should that matter in this case, -1==-1 always evaluates to true, and myobj.nil? evaluates to false. so the net result should always be false.
Shekhar
2010-05-14 06:36:24
oh ok, so if it was used in conditional expression likeif (-1 == -1) and (myobj.nil?)both, would have same result?
Shekhar
2010-05-14 06:42:23
Jörg W Mittag
2010-05-14 06:48:15
Jorg, thanks. this is really helpful, as I never realized print was being considered part of the expression.
Shekhar
2010-05-14 07:10:37
+3
A:
&& and and have different precedence. In particular, and has very low precedence, lower than almost anything else.
The return value of print is nil, at least in Ruby 1.8. Therefore, the first expression reduces like this:
(print (-1 == -1)) and myobj.nil?
(print true) and myobj.nil? # => with the side-effect of printing "true"
nil and myobj.nil?
nil
While the second expression reduces like this:
print ((-1 == -1) && myobj.nil?)
print ( true && myobj.nil?)
print myobj.nil?
print false # => with the side-effect of printing "false"
nil
Jörg W Mittag
2010-05-14 06:53:34