将List的元素作为参数传递给具有可变参数的函数 [英] Passing elements of a List as parameters to a function with variable arguments
问题描述
例如,我有一个名为或
的函数,它被定义为;
或(filters:FilterDefinition *)
然后我有一个列表:
列表(X,Y,Z)
我现在需要做的是调用或
,比如
或(func(X),func(Y),func(Z))
和预期的一样,列表的长度可能会改变。
在Scala中执行此操作的最佳方式是什么?
解决方案
code> def printme(s:String *)= s.foreach(println)
scala> printme(List(a,b,c))
< console>:9:error:type mismatch;
found:List [String]
required:String
printme(List(a,b,c))
你真的需要用解开列表参数:_ *
操作符
scala> val mylist = List(1,2,3)
scala> printme(mylist:_ *)
1
2
3
I have a function called or
for example, which is defined as;
or(filters: FilterDefinition*)
And then I have a list:
List(X, Y, Z)
What I now need to do is call or
like
or(func(X), func(Y), func(Z))
And as expected the length of the list may change.
What's the best way to do this in Scala?
Take a look at this example, I will define a function printme that takes vargs of type String
def printme(s: String*) = s.foreach(println)
scala> printme(List("a","b","c"))
<console>:9: error: type mismatch;
found : List[String]
required: String
printme(List(a,b,c))
What you really need to un-pack the list into arguments with the :_*
operator
scala> val mylist = List("1","2","3")
scala> printme(mylist:_*)
1
2
3
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