在 Scala 中执行块 n 次是否有简短的语法？
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Is there a brief syntax for executing a block n times in Scala?
提问by Craig P. Motlin
I find myself writing code like this when I want to repeat some execution n times:
当我想重复执行 n 次时，我发现自己编写了这样的代码：
for (i <- 1 to n) { doSomething() }
I'm looking for a shorter syntax like this:
我正在寻找更短的语法，如下所示：
n.times(doSomething())
Does something like this exist in Scala already?
Scala 中已经存在这样的东西了吗？
EDIT
编辑
I thought about using Range's foreach() method, but then the block needs to take a parameter which it never uses.
我想过使用 Range 的 foreach() 方法，但是该块需要采用一个它从不使用的参数。
(1 to n).foreach(ignored => doSomething())
采纳答案by missingfaktor
You could easily define one using Pimp My Library pattern.
您可以使用 Pimp My Library 模式轻松定义一个。
scala> implicit def intWithTimes(n: Int) = new {
| def times(f: => Unit) = 1 to n foreach {_ => f}
| }
intWithTimes: (n: Int)java.lang.Object{def times(f: => Unit): Unit}
scala> 5 times {
| println("Hello World")
| }
Hello World
Hello World
Hello World
Hello World
Hello World
回答by sblundy
The Range class has a foreach method on it that I think is just what you need. For example, this:
Range 类有一个 foreach 方法，我认为这正是您所需要的。例如，这个：
0.to(5).foreach(println(_))
produced
产生
0
1
2
3
4
5
0
1
2
3
4
5
回答by Apocalisp
With scalaz 5:
使用scalaz 5：
doSomething.replicateM[List](n)
With scalaz 6:
使用scalaz 6：
n times doSomething
And that works as you would expect with most types (more precisely, for every monoid):
这与您对大多数类型的期望一样（更准确地说，对于每个 monoid）：
scala> import scalaz._; import Scalaz._; import effects._;
import scalaz._
import Scalaz._
import effects._
scala> 5 times "foo"
res0: java.lang.String = foofoofoofoofoo
scala> 5 times List(1,2)
res1: List[Int] = List(1, 2, 1, 2, 1, 2, 1, 2, 1, 2)
scala> 5 times 10
res2: Int = 50
scala> 5 times ((x: Int) => x + 1).endo
res3: scalaz.Endo[Int] = <function1>
scala> res3(10)
res4: Int = 15
scala> 5 times putStrLn("Hello, World!")
res5: scalaz.effects.IO[Unit] = scalaz.effects.IO$$anon@36659c23
scala> res5.unsafePerformIO
Hello, World!
Hello, World!
Hello, World!
Hello, World!
Hello, World!
You could also say doSomething replicateM_ 5which only works if your doSomethingis an idiomatic value (see Applicative). It has better type-safety, since you can do this:
您也可以说doSomething replicateM_ 5这仅在您doSomething是惯用值时才有效（请参阅 参考资料Applicative）。它具有更好的类型安全性，因为您可以这样做：
scala> putStrLn("Foo") replicateM_ 5
res6: scalaz.effects.IO[Unit] = scalaz.effects.IO$$anon@8fe8ee7
but not this:
但不是这个：
scala> { System.exit(0) } replicateM_ 5
<console>:15: error: value replicateM_ is not a member of Unit
Let me see you pull that off in Ruby.
让我看看你在 Ruby 中实现了这一点。
回答by huynhjl
I'm not aware of anything in the library. You can define a utility implicit conversion and class that you can import as needed.
我不知道图书馆里有什么。您可以定义可以根据需要导入的实用程序隐式转换和类。
class TimesRepeat(n:Int) {
def timesRepeat(block: => Unit): Unit = (1 to n) foreach { i => block }
}
object TimesRepeat {
implicit def toTimesRepeat(n:Int) = new TimesRepeat(n)
}
import TimesRepeat._
3.timesRepeat(println("foo"))
Rahul just posted a similar answer while I was writing this...
在我写这篇文章的时候，拉胡尔刚刚发布了一个类似的答案......
回答by Adrian
It can be as simple as this:
它可以像这样简单：
scala> def times(n:Int)( code: => Unit ) {
for (i <- 1 to n) code
}
times: (n: Int)(code: => Unit)Unit
scala> times(5) {println("here")}
here
here
here
here
here
回答by Less
def times(f: => Unit)(cnt:Int) :Unit = {
List.fill(cnt){f}
}
