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Lazy pattern matching in Scala made even lazier with scalaz's Need.
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| Welcome to Scala version 2.9.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_05). | |
| Type in expressions to have them evaluated. | |
| Type :help for more information. | |
| scala> import scalaz._ | |
| import scalaz._ | |
| scala> val (foo, bar, baz) = (Need({println("foo") ; "foo"}), Need({ println("bar") ; "bar"}), Need({ println("baz") ; "baz"})) | |
| foo: scalaz.Need[java.lang.String] = scalaz.Need$$anon$4@1d90d034 | |
| bar: scalaz.Need[java.lang.String] = scalaz.Need$$anon$4@e551516 | |
| baz: scalaz.Need[java.lang.String] = scalaz.Need$$anon$4@73db5eb3 | |
| scala> lazy val List(a, b, c) = List(foo, bar, baz) | |
| a: scalaz.Need[java.lang.String] = <lazy> | |
| b: scalaz.Need[java.lang.String] = <lazy> | |
| c: scalaz.Need[java.lang.String] = <lazy> | |
| scala> a | |
| res0: scalaz.Need[java.lang.String] = scalaz.Need$$anon$4@1d90d034 | |
| scala> b | |
| res1: scalaz.Need[java.lang.String] = scalaz.Need$$anon$4@e551516 | |
| scala> c | |
| res2: scalaz.Need[java.lang.String] = scalaz.Need$$anon$4@73db5eb3 | |
| scala> a.value | |
| foo | |
| res3: java.lang.String = foo | |
| scala> b.value | |
| bar | |
| res4: java.lang.String = bar | |
| scala> c.value | |
| baz | |
| res5: java.lang.String = baz |
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What does the first level of laziness (lazy vals) gain you that using Need doesn't?