Re: One out of three ain't bad (order)

Avoid the obfuscating punctuation:

if( 1 == grep $_, $x, $y, $z )
[download]

Or, using a more versatile trick:

if( 1 == !!$x + !!$y + !!$z )
[download]

But note that some will kvetch that the fact of !0 == 1 is "not defined" in Perl. But that ship has already sailed; practicality actually prevents Perl from changing the value of "true" to something other than 1, even if this property was intentionally left undefined.

- tye

Comment on Re: One out of three ain't bad (order) Select or Download Code

Replies are listed 'Best First'.
Re^2: One out of three ain't bad (order) by ph713 (Pilgrim) on Oct 22, 2005 at 08:42 UTC
I like this !! based solution the best. It's the most logical way to solve the problem - convert the arguments to boolean truth values, then sum them. You could do it for an arbitrarily long list of variables with: `use List::Util qw( sum ); if( 1 == sum( map { !!$_ } @inputs) ) { ... }` [download]	[reply] [d/l]
Re^3: One out of three ain't bad (order) by Tanktalus (Canon) on Oct 22, 2005 at 14:46 UTC
As long as you're going to go through the same code path for each element in the array anyway, why not de-obfuscate that a bit and use the ternary operator? `if (1 == sum( map { $_ ? 1 : 0 } @inputs ))` [download] While tye may be right that, in practice, Perl can't change the value of !0, I just don't like relying on obscure implementation details when clear, well-defined implementation details are available to me to accomplish the same thing in a readable, understandable, and maintainable manner.	[reply] [d/l]
Re^4: One out of three ain't bad (order) by eric256 (Parson) on Oct 23, 2005 at 04:02 UTC
Why are you using sum instead of grep here? If you are summing 1 and 0 then the sum will always be the same as the grep answer. For what it's worth I woul disagree that ($_ ? 1 : 0) is clearer in anyway than !!$_. To me the !! version yells bool, while I had to read the ?: version twice before comprehension came. That might be in part because Perl6's new bool context (don't shoot me if that isn't the right wording) lets you ask for a true/false value explicity so thats what I would expext !! to do. ___________ Eric Hodges	[reply]
Re^5: One out of three ain't bad (order) by Tanktalus (Canon) on Oct 23, 2005 at 15:39 UTC
Re^6: One out of three ain't bad (order) by sauoq (Abbot) on Oct 26, 2005 at 04:39 UTC
Re^6: One out of three ain't bad (order) by eric256 (Parson) on Oct 24, 2005 at 02:08 UTC
Re^2: One out of three ain't bad (order) by Dr. Mu (Hermit) on Oct 23, 2005 at 04:21 UTC
Or, shorter: `if (2 == !$x + !$y + !$z) {`	[reply] [d/l]
Re^3: One out of three ain't bad (order) by mrborisguy (Hermit) on Oct 24, 2005 at 04:51 UTC
Which seems to work right now, since you've seen the progression of what came before. But in 2 months, I can imagine you thinking, "Now why in the world is there a two in there... I must want two of them to be true" on your first look while skimming your code. -Bryan	[reply]
Re^4: One out of three ain't bad (order) by Dr. Mu (Hermit) on Oct 24, 2005 at 17:23 UTC
True enough. I suppose what's lacking is a commonly-accepted idiom for Boolean conversions. We have `0+` for numeric conversions; so perhaps `!!` wasn't so bad after all, even though two operations are implied (unless the compiler optimizes them out). `0\|\|` and `1&&` would also work.	[reply]
Re^5: One out of three ain't bad (! !!) by tye (Sage) on Oct 25, 2005 at 03:37 UTC
Re^6: One out of three ain't bad (! !!) by Dr. Mu (Hermit) on Oct 26, 2005 at 01:59 UTC
Some notes below your chosen depth have not been shown here

In Section Seekers of Perl Wisdom