Help with while condition OR condition

Ruby
1

Hi everyone,

This is my first post, so I hope I don’t sound too inexperienced…

I’m trying to teach myself Ruby, and have run into an issue with a while
statement that will break if an input is “exit” or “quit”.
As of right now, it works if exit is input, but not quit

I know I am completely misusing the entire thing, but here is what I
came up with:

EXIT = “exit” #need constants since Ruby gets pissed at string literals
QUIT = “quit” #in a comparison

print "Input: "
input = gets
while input.chomp.downcase != (EXIT || QUIT) #only works for exit

#Do something

print "Input: " #pick up the next input and check it
input = gets
end

I know that Ruby has a lot of shortcuts, but if you post any please
explain how they work (or provide a link to a good explanation.

Thanks!

2

Don’t do that, just use eql?

begin
print "input: "
input = gets.chomp.downcase
end while not (input.eql? “exit” or input.eql? “quit”)

3

On Saturday, May 7, 2011, Bill W. [email protected] wrote:

EXIT = “exit” #need constants since Ruby gets pissed at string literals
QUIT = “quit” #in a comparison

print "Input: "
input = gets
while input.chomp.downcase != (EXIT || QUIT) #only works for exit

The || operator returns the operand on the left if it is “true-ish”
(anything other than nil or false), otherwise it evaluates and returns
the operand on the right. So (EXIT || QUIT) where EXIT=“exit” just
evaluates to “exit”.

You could do this with the || between two comparisons (rather than two
options in one comparison).

Also, Ruby has no problem with comparisons against string literals;
what made you think it did?

4

Oops, small typo:

while command != "exit" || command != "quit" ...

should be:

while command != "exit" && command != "quit" ...

–
John F.
Principal Consultant, BitsBuilder
LI: http://www.linkedin.com/in/johnxf
SO: User John Feminella - Stack Overflow

5

This line doesn’t behave in the way I think you expect:

while input.chomp.downcase != (EXIT || QUIT) #only works for exit

This says, “check if the downcased input is not equal to the value of
the expression EXIT || QUIT”. What is the value of that expression?
In this case, it will resolve to EXIT, since the string “exit” is not
false or nil and is thus true. The value of QUIT is never evaluated.
So, if the input is not equal to exit, the while loop continues.

Why is that? In Ruby, all expressions have both a “value” and a
“truthiness”. An expression is “falsy” if it evaluates to either
false or nil; otherwise it is truthy. In the case of an
expression like foo || bar, the truth table would look like this:

  • foo is truthy, bar is truthy: result is foo and truthy
  • foo is truthy, bar is falsy: result is foo and truthy
  • foo is falsy, bar is truthy: result is bar and truthy
  • foo is falsy, bar is falsy: result is bar and falsy

So you can see that in your case, foo is EXIT and bar is QUIT, both of
which are truthy values; thus the expression is “exit”. To get what
you what want, try something like this:

command = input.chomp.downcase
while command != "exit" || command != "quit" ...

Or, more succinctly:

case command
when "exit", "quit"
  puts "exiting!"; ...
else
  # do stuff
end

~ jf

John F.
Principal Consultant, BitsBuilder
LI: http://www.linkedin.com/in/johnxf
SO: User John Feminella - Stack Overflow

6

On 05/07/2011 03:37 PM, Bill W. wrote:

EXIT = “exit” #need constants since Ruby gets pissed at string literals
QUIT = “quit” #in a comparison

print "Input: "
input = gets
while input.chomp.downcase != (EXIT || QUIT) #only works for exit

It can help to take apart expressions in irb (interactive ruby):

$ irb

EXIT = “exit”
=> “exit”

QUIT = “quit”
=> “quit”

EXIT || QUIT
=> “exit”

“quit” == (EXIT || QUIT)
=> false

#Do something

print "Input: " #pick up the next input and check it
input = gets
end

I know that Ruby has a lot of shortcuts, but if you post any please
explain how they work (or provide a link to a good explanation.

Something to tinker with:

print "Input: "
while input = gets
case s = input.chomp.downcase
when “exit”, “quit”
puts “You wanted to #{s} this mighty fine program?”
break
else
puts “Why do you say ‘#{s}’?”
print "Input: "
end
end
puts “Done.”

7

(EXIT || QUIT) will always return “exit” as EXIT is never nil or
false, so in your case, the input string is never checked against
QUIT.

Try this:

 print "Input: "
 input = gets

 until ["exit", "quit"].include? input.chomp.downcase
   # Do something
   print "Input: "
   input = gets
 end

As you can see you can do it without constants. the method include?
checks if an element exists in the array. until is basically an
inverted while.

8

On Sat, May 7, 2011 at 5:37 PM, Bill W. [email protected] wrote:

EXIT = “exit” #need constants since Ruby gets pissed at string literals
QUIT = “quit” #in a comparison

It doesn’t get pissed at string literals in comparison, its just that it
was
able to tell that the way you were comparing didn’t make sense (because
string literal is known at time you wrote it, not dynamically looked up,
so
the comparison is also known, and doesn’t make sens). So it was warning
you
of the issue. It didn’t complain when you stored then in vars/constants,
because it doesn’t know their value until runtime. They could
hypothetically
be false or nil, so it isn’t conspicuously an error.

e.g.

if “some literal”

do something

end # !> string literal in condition

if “some literal” || “other”

do something

end # !> string literal in condition

9

Wow! I would say this is the most replies I have ever had on a
programming topic!

Thanks for the in-depth description of how || works, I was WAY off.

I tried this with until already once, but I had the rest of it wrong and
it failed.

command = input.chomp.downcase
until [“exit”, “quit”].include? input.chomp.downcase
are both exactly the type of shortcuts i would have never though of!

Thanks again!

10

Everyone should use ‘and’ and ‘or’ by default instead of && and ||.
Code reads better that way. Only if you have a specific reason to,
should you use && or ||.

It’s not a good idea to make a blanket rule like that, imo. The “and”
keyword is not really a substitute for “&&”, since it has different
precedence. It’s best viewed as a control flow modifier (like “if” or
“unless” when at the end of an expression), rather than a true logical
operator.

If you don’t know that it’s not quite the same, this can get you into
big trouble by leading to subtle bugs. Consider this code, for
instance:

missiles_armed = true
=> true
go_for_launch = false
=> false

Using &&

go_for_launch && missiles_armed ? :fire_ze_missiles : :abort_launch
=> :abort_launch # Looks good here.

Using “and”

go_for_launch and missiles_armed ? :fire_ze_missiles : :abort_launch
=> false # Uh-oh! We didn’t get the :abort_launch we were
expecting…

~ jf

John F.
Principal Consultant, BitsBuilder
LI: http://www.linkedin.com/in/johnxf
SO: User John Feminella - Stack Overflow

11

Hi,

A lot of beginners make the same mistake you did. ‘Compound
conditionals’ have to be written like separate conditionals and then
hooked together with an ‘or’ or ‘and’. For instance if you wanted to do
something only if a number were greater than 5 and less then 10, you
would do this:

x > 5
x < 10
and

if x > 5 and x < 10
#do something
end

You don’t write it like this:

if x > 5 and < 10
#do something
end

python (the language you should have chosen to learn) has the handy:

if 5 < x < 10:
#do something

In ruby, everyone should use ‘and’ and ‘or’ by default instead of && and
||. Code reads better that way. Only if you have a specific purpose in
mind should you use && or ||.

Good luck.

12

7stud – wrote in post #997311:

In ruby, everyone should use ‘and’ and ‘or’ by default instead of && and
||.

I would advise exactly the opposite: there are many traps for the unwary
if you use ‘and’ and ‘or’. Two prime examples:

val = 10
=> 10

ok = val < 3 or val > 5
=> true

ok
=> false

a = true
=> true

b = not a
SyntaxError: compile error
(irb):7: syntax error, unexpected kNOT

Use ‘||’ and ‘!’ respectively and you won’t have a problem.

13

Brian C. wrote in post #997379:

Use ‘||’ and ‘!’ respectively and you won’t have a problem.

lol.

14

On Mon, May 9, 2011 at 10:11 PM, 7stud – [email protected]
wrote:

Brian C. wrote in post #997379:

Use ‘||’ and ‘!’ respectively and you won’t have a problem.

lol.

Should’ve tried the code first:

irb(main):001:0> val = 10
=> 10
irb(main):002:0> ok = val < 3 || val > 5
=> true
irb(main):003:0> ok
=> true
irb(main):004:0> a = true
=> true
irb(main):005:0> b = !a
=> false

Using only “or” or “not” obviously doesn’t lead to the expected
results, while using || and ! do.

–
Phillip G.

Though the folk I have met,
(Ah, how soon!) they forget
When I’ve moved on to some other place,
There may be one or two,
When I’ve played and passed through,
Who’ll remember my song or my face.