scala Int vs Integer：类型不匹配，找到：Int，需要：String
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Int vs Integer: type mismatch, found: Int, required: String
提问by pcjuzer
I type these to the scala interpreter:
我将这些输入到 Scala 解释器中：
val a : Integer = 1;
val b : Integer = a + 1;
And I get the message:
我收到消息：
<console>:5: error: type mismatch;
found : Int(1)
required: String
val b : Integer = a +1
^
Why? How can I solve this? This time I need Integers due to Java interoperability reasons.
为什么？我该如何解决这个问题？这次由于 Java 互操作性原因我需要整数。
回答by Flaviu Cipcigan
This question is almost a duplicate of: Scala can't multiply java Doubles?- you can look at my answeras well, as the idea is similar.
这个问题几乎是重复的：Scala 不能乘以 java Doubles？- 你也可以看看我的答案，因为想法是相似的。
As Eastsunalready hinted, the answer is an implicit conversionfrom an java.lang.Integer(basically a boxed intprimitive) to a scala.Int, which is the Scala way of representing JVM primitive integers.
正如Eastsun已经暗示的那样，答案是从 an （基本上是盒装原语）到 a的隐式转换，这是表示 JVM 原语整数的 Scala 方式。java.lang.Integerintscala.Int
implicit def javaToScalaInt(d: java.lang.Integer) = d.intValue
And interoperability has been achieved - the code snipped you've given should compile just fine! And code that uses scala.Intwhere java.lang.Integeris needed seems to work just fine due to autoboxing. So the following works:
并且已经实现了互操作性 - 您提供的代码应该可以很好地编译！由于自动装箱，scala.Int在java.lang.Integer需要的地方使用的代码似乎工作得很好。所以以下工作：
def foo(d: java.lang.Integer) = println(d)
val z: scala.Int = 1
foo(z)
Also, as michaelkebesaid, do not use the Integertype - which is actually shorthand for scala.Predef.Integeras it is deprecated and most probably is going to be removed in Scala 2.8.
此外，正如michaelkebe所说，不要使用Integer类型 - 这实际上是简写，scala.Predef.Integer因为它已被弃用，并且很可能将在 Scala 2.8 中删除。
EDIT: Oops... forgot to answer the why. The error you get is probably that the scala.Predef.Integertried to mimic Java's syntactic sugar where a + "my String"means string concatenation, ais an int. Therefore the +method in the scala.Predef.Integertype only does string concatenation (expecting a Stringtype) and no natural integer addition.
编辑：哎呀...忘了回答为什么。你得到的错误可能是scala.Predef.Integer试图模仿 Java 的语法糖，其中a + "my String"意味着字符串连接，a是一个int. 因此类型中的+方法scala.Predef.Integer只进行字符串连接（期望String类型），没有自然整数加法。
-- Flaviu Cipcigan
——弗拉维乌·西普西根
回答by Eastsun
Welcome to Scala version 2.7.3.final (Java HotSpot(TM) Client VM, Java 1.6.0_16).
Type in expressions to have them evaluated.
Type :help for more information.
scala> implicit def javaIntToScala(n: java.lang.Integer) = n.intValue
javaIntToScala: (java.lang.Integer)Int
scala> val a: java.lang.Integer = 1
a: java.lang.Integer = 1
scala> val b: java.lang.Integer = a + 1
b: java.lang.Integer = 2
回答by michael.kebe
First of all you should use java.lang.Integerinstead of Integer.
首先，您应该使用java.lang.Integer而不是Integer.
Currently I don't know why the error occurs.
目前我不知道为什么会发生错误。
ais an instance of java.lang.Integerand this type doesn't have a method named +. Furthermore there is no implicit conversion to Int.
a是 的实例，java.lang.Integer并且此类型没有名为 的方法+。此外，没有隐式转换为Int.
To solve this you can try this:
要解决此问题，您可以尝试以下操作：
scala> val a: java.lang.Integer = 1 a: java.lang.Integer = 1 scala> val b: java.lang.Integer = a.intValue + 1 b: java.lang.Integer = 2
