接受的答案实际上是错误的。
您没有获得 Map.+= 元组的原因是该方法被带有两个或更多参数的第二个方法重载。
仅当参数数量错误时，编译器才会尝试元组。但是，如果你给它两个参数，并且有一个方法需要两个参数，那么它就会选择它，即使它无法进行类型检查。
在某些东西起作用之前，它不会开始尝试所有可能的组合，因为这会充满晦涩难懂。 （参见隐式。）
scala> def f(p: Pair[String, Int]) = { val (s,i) = p; s.toInt + i }
f: (p: Pair[String,Int])Int
scala> f("2",3) // ok to tuple
res0: Int = 5
scala> trait F { def +=(p: Pair[String, Int]) = f(p) }
defined trait F
scala> val foo = new F {}
foo: F = $anon$1@6bc77f62
scala> foo += ("2",3) // ok to tuple
res1: Int = 5
scala> trait G { def +=(p: Pair[String, Int]) = f(p); def +=(p:(String,Int),q:(String,Int),r:(String,Int)*) = f(p)+f(q)+(r map f).sum }
defined trait G
scala> val goo = new G {}
goo: G = $anon$1@183aeac3
scala> goo += ("2",3) // sorry
<console>:12: error: type mismatch;
found : String("2")
required: (String, Int)
goo += ("2",3)
^
scala> goo += (("2",3),("4",5),("6",7))
res3: Int = 27
如果我不提你的朋友和我的朋友 -Xlint，那就太失职了，它会警告不良的参数调整：
apm@mara:~/tmp$ skala -Xlint
Welcome to Scala version 2.11.0-20130811-132927-95a4d6e987 (OpenJDK 64-Bit Server VM, Java 1.7.0_25).
Type in expressions to have them evaluated.
Type :help for more information.
scala> def f(p: Pair[String, Int]) = { val (s,i) = p; s.toInt + i }
f: (p: Pair[String,Int])Int
scala> f("2",3)
<console>:9: warning: Adapting argument list by creating a 2-tuple: this may not be what you want.
signature: f(p: Pair[String,Int]): Int
given arguments: "2", 3
after adaptation: f(("2", 3): (String, Int))
f("2",3)
^
res0: Int = 5
scala> List(1).toSet()
<console>:8: warning: Adapting argument list by inserting (): this is unlikely to be what you want.
signature: GenSetLike.apply(elem: A): Boolean
given arguments: <none>
after adaptation: GenSetLike((): Unit)
List(1).toSet()
^
res3: Boolean = false
关于适应的危险，请参阅自适应推理益智游戏 http://scalapuzzlers.com/#pzzlr-036 and 这个新的很常见 http://scalapuzzlers.com/#pzzlr-040因为我们知道括号的存在或不存在很大程度上是风格问题，当括号确实很重要时，错误地使用它们会导致类型错误。
在存在重载的情况下调整元组：
scala> class Foo {
| def f[A](a: A) = 1 // A can be (Int,Int,Int)
| def f[A](a: A, a2: A) = 2
| }
defined class Foo
scala> val foo = new Foo
foo: Foo = Foo@2645d22d
scala> foo.f(0,0,0)
<console>:10: warning: Adapting argument list by creating a 3-tuple: this may not be what you want.
signature: Foo.f[A](a: A): Int
given arguments: 0, 0, 0
after adaptation: Foo.f((0, 0, 0): (Int, Int, Int))
foo.f(0,0,0)
^
res9: Int = 1
