-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1
Maybe I misunderstood you, but you aren't quite right. In third solution, *list comprehension* will produce all the pairs. But you're applying 'null' to it, and since everything in Haskell is lazy by default, only *one* pair will be produced.
Good point! Lovely way to show this is with 'trace':
:m +Debug.Trace null [x | x <- [trace "a" 5, trace "b" 10, trace "c" 15]] False null [x | x <- [trace "a" 5, trace "b" 10, trace "c" 15], x > 0] a False null [x | x <- [trace "a" 5, trace "b" 10, trace "c" 15], x > 5] a b False null [x | x <- [trace "a" 5, trace "b" 10, trace "c" 15], x > 10] a b c False
(Of course, this is less beautifully illustrated with 'length', because it doesn't actually evaluate the values ...:
length [trace "a" 5, trace "b" 10, trace "c" 15] == 0 False
-----BEGIN PGP SIGNATURE----- Version: GnuPG v2.0.17 (MingW32) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org/ iQIcBAEBAgAGBQJOL2eHAAoJEDiWqExGnQ/QIyMQAK5UhXziU8ctse3K2aUrwrbm BrQd4ec7i3nCqM0ykkwV9jAAZdHKL3Hu1gWPzH7kBOmcSBcfE3/0woZOna10cSry jovsAHt5eiE/ujqGWkWeidVSu3wcNcoMXm0ZjdTG5FPx9O+YXvmCRzaxUp0lYZzO w3tRy4tQPLogcfR72HLlk6IBdG/8d04shQ8lTgRmbRKALI/VN2N8+qqYSsZySiF9 SmE2yQIG3pwSm6QNCCeCGudoz2qjN7SIM4443pjOMJODda1dLpliuC0eo9q+4YSW ZOH8oJFGembav2C6epYw8qLkSB0bvJltohspZK5GifeSUEWZT36WAbqF0NOAOn6D du0JW/bEazLiph2NNC2qzqjWDAnoYmY1Nqx3tob2UDMslf8h2H8vMvTiUKKU2tED h33HgLWfW0l+MPaAvIdTDNscm1/dExleyrtsc6PMA5If650wdjITzHFqQ+9fCYD8 Qbd475cOL5ZkbQjGRlIzgq5goH9JUOv5wqIjz9LOKd1q6KGLq5UN3OTEyjkDt1vo zGtMjQqgha/M3sznXVUTkBc12gFotVYuUmUCOQFEafjxhtN46JR0fNHJjOFzpqb6 Jkqd75GvOHJCfNUkMc42ZDEWXDzBO38f3Vgouf6VRK+/yDqqOSXyK+/SspQwLuFk Ph9fI7dGT178rNQpdU3A =gZUe -----END PGP SIGNATURE-----