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comp.lang.ruby

Line breaking in Ruby can be dangerous

Sky Yin

1/23/2006 12:40:00 PM

After programing in ruby for the past 3 months, I must say I'm
happier. However, I find where to break a long expression into lines
can be potentially dangerous and very hidden to debug. For example:

>>begin
?> (1 * 2
?> + 1)
?> end
=> 1

I guess this may be attributed to the Ruby convention that the return
value by default is from the last line. The weird thing is that even
parentheses can't guarantee the result to be expected.

I'm still using 1.8.2 on win32. Is it a bug of that version?
9 Answers

Florian Frank

1/23/2006 1:13:00 PM

0

On 2006-01-23 21:39:33 +0900, Sky Yin wrote:
> >>begin
> ?> (1 * 2
> ?> + 1)
> ?> end
> => 1
>
> I guess this may be attributed to the Ruby convention that the return
> value by default is from the last line. The weird thing is that even
> parentheses can't guarantee the result to be expected.

"1 * 2" and "+ 1" are both valid ruby expressions, the latter "(unary +)
1". How should Ruby find out, that you wanted to continue your first
expression in the next line?

If you want to avoid this, only break lines after operators, ",", etc.
like

1 * 2 +
1

You can also use

1 * 2 + 1

but I try to avoid this way, I think it's kind of ugly.

--
Florian Frank

Daniel Schüle

1/23/2006 1:43:00 PM

0

>> I guess this may be attributed to the Ruby convention that the return
>> value by default is from the last line. The weird thing is that even
>> parentheses can't guarantee the result to be expected.
>
> "1 * 2" and "+ 1" are both valid ruby expressions, the latter "(unary +)
> 1". How should Ruby find out, that you wanted to continue your first
> expression in the next line?

because of (

I would also expect either a SyntaxError or correct result (3)

the latter seems more natural to me

Regards, Daniel


Adriano Ferreira

1/23/2006 1:46:00 PM

0

On 1/23/06, Florian Frank <flori@nixe.ping.de> wrote:
> "1 * 2" and "+ 1" are both valid ruby expressions, the latter "(unary +)
> 1". How should Ruby find out, that you wanted to continue your first
> expression in the next line?

Ruby would find it out because of the leading "(" or that would be
what I and apparently Sky Yin expect.

This works:
irb(main):001:0> (1
irb(main):002:1> )
=> 1

This too:
irb(main):003:0> begin (1
irb(main):004:2> ) end
=> 1

But I think I understand why it works this way. In Yin's example:

(1*2
+1)

is interpreted just like
(1*2;
+1)

what would not happen if he had written
(1*2+
1)

which gives a proper answer.

Kind regards,
Adriano.

Christian Neukirchen

1/23/2006 2:03:00 PM

0

Sky Yin <sky.yin@gmail.com> writes:

>>>begin
> ?> (1 * 2
> ?> + 1)
> ?> end
> => 1

Simply use:
1 * 2 +
1

or:

1 * 2 + 1

> I guess this may be attributed to the Ruby convention that the return
> value by default is from the last line. The weird thing is that even
> parentheses can't guarantee the result to be expected.

That's because you can have multiple statements in (), e.g.:

(puts "foo"
puts "bar")

#=> foo
#=> bar
--
Christian Neukirchen <chneukirchen@gmail.com> http://chneuk...

Sky Yin

1/23/2006 2:19:00 PM

0

Thanks. When I first found it out, it's really a *big* surprise. This
actually took me quite a while to debug, cause the long expression
itself that I wrote simply looked nothing-wrong at the first glance.


On 1/23/06, Adriano Ferreira <a.r.ferreira@gmail.com> wrote:
> On 1/23/06, Florian Frank <flori@nixe.ping.de> wrote:
> > "1 * 2" and "+ 1" are both valid ruby expressions, the latter "(unary +)
> > 1". How should Ruby find out, that you wanted to continue your first
> > expression in the next line?
>
> Ruby would find it out because of the leading "(" or that would be
> what I and apparently Sky Yin expect.
>
> This works:
> irb(main):001:0> (1
> irb(main):002:1> )
> => 1
>
> This too:
> irb(main):003:0> begin (1
> irb(main):004:2> ) end
> => 1
>
> But I think I understand why it works this way. In Yin's example:
>
> (1*2
> +1)
>
> is interpreted just like
> (1*2;
> +1)
>
> what would not happen if he had written
> (1*2+
> 1)
>
> which gives a proper answer.
>
> Kind regards,
> Adriano.
>
>


--
Blog >>> http://spaces.msn.com/members/s...

Christian Neukirchen

1/23/2006 4:42:00 PM

0

Sky Yin <sky.yin@gmail.com> writes:

> Thanks. When I first found it out, it's really a *big* surprise. This
> actually took me quite a while to debug, cause the long expression
> itself that I wrote simply looked nothing-wrong at the first glance.

I recommend an intelligent-indenting editor.
--
Christian Neukirchen <chneukirchen@gmail.com> http://chneuk...

MenTaLguY

1/23/2006 10:14:00 PM

0

Quoting Christian Neukirchen <chneukirchen@gmail.com>:

> That's because you can have multiple statements in (), e.g.:
>
> (puts "foo"
> puts "bar")
>
> #=> foo
> #=> bar

Whoa.

-mental

MenTaLguY

1/23/2006 10:44:00 PM

0

Quoting Florian Frank <flori@nixe.ping.de>:

> "1 * 2" and "+ 1" are both valid ruby expressions, the latter
> "(unary +)
> 1". How should Ruby find out, that you wanted to continue your
> first expression in the next line?

The parenthesis. It shouldn't parse if they were treated as
separate expressions.

-mental

MenTaLguY

1/23/2006 11:01:00 PM

0

Quoting mental@rydia.net:

> Quoting Florian Frank <flori@nixe.ping.de>:
>
> > "1 * 2" and "+ 1" are both valid ruby expressions, the latter
> > "(unary +)
> > 1". How should Ruby find out, that you wanted to continue your
> > first expression in the next line?
>
> The parenthesis. It shouldn't parse if they were treated as
> separate expressions.

Okay, never mind. Now I know the real answer.

(1 * 2
+ 1)

parses just like:

(1 * 2;+ 1)

-mental