blocks|key|704114|text|getOrElse的签名是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|704115|def+getOrElse[B1+>:+B](key:+A,+default:+⇒+B1):+B1+|code-block|syntax|javascript|704116|和你的声明|704117|val+d:+Double+=+m.getOrElse('c,+0)|704118|您正在告诉Scala+B1应该是Double。但是Double不是Int的超级类型，通常的超级类型是AnyVal.所以你不能做那个任务..。|704119|尝试将Int结果转换为Double：|704120|val+d+/*+:+Double+*/+=+m.getOrElse('c,+0).toDouble|704121|entityMap^0|0|9|0|0|0|0|B|2|G|6|P|6|X|3|1E|6|0|3|3|B|6|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1J|8|@]|D|@]|E|$]]]|V|$]]

The signature of <code>getOrElse</code> is:

<pre><code>def getOrElse[B1 &gt;: B](key: A, default: ⇒ B1): B1 
</code></pre>

With your statement

<pre><code>val d: Double = m.getOrElse('c, 0)
</code></pre>

You are telling Scala that <code>B1</code> ought to be <code>Double</code>. But <code>Double</code> is not a super-type of <code>Int</code>, the common supertype instead is <code>AnyVal</code>. Therefore, you cannot make that assignment...

Try casting the <code>Int</code> result to a <code>Double</code> instead:

<pre><code>val d /* : Double */ = m.getOrElse('c, 0).toDouble
</code></pre>

blocks|key|495398|text|因为map在其第二类参数中是协变的，所以我们需要使用特殊的技巧来避免反方差问题，即为这种类型指定一个上限：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|495399|def+getOrElse[B1+>:+B](key:+A,+default:+=>+B1):+B1|code-block|syntax|javascript|495400|因为您显式地提供返回类型为Double，所以scalac将其常见的超级类型解析为AnyVal，因为Int和Double是AnyVal的子类型。在第二个例子中，scalac看到了类型归属，因此它正确地推断了B1，然后将Int转换为Double。|offset|length|style|CODE|495401|entityMap^0|0|0|D|6|14|6|1D|3|1H|6|1O|6|2U|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|N|$]]

Because map is covariant in its second type parameter, we need to use special trick to avoid contravariance problem, i.e. specify an upper bound for this type:

<pre><code>def getOrElse[B1 &gt;: B](key: A, default: =&gt; B1): B1
</code></pre>

Because you explicitly provide that the return type is <code>Double</code>, scalac resolves its common supertype as <code>AnyVal</code>, because <code>Int</code> and <code>Double</code> are subtypes of <code>AnyVal</code>. In your second case, scalac sees a type ascription, so it correctly infers <code>B1</code> and then casts Int to Double.

blocks|key|495418|text|我正在尝试更好地阅读-Ytyper-debug输出。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|495419|对于预期的类型Double，0类型被视为Double，0.0。那是因为人们讨厌不断扩大的转变。|495420|这是在为B1求解之前。B1被认为是AnyVal，因为它是A/Int和Double的lub。|495421|%7C++++%7C--+Double+TYPEmode+(site:+value+d++in+$iw)+
%7C++++%7C++++\->+Double
%7C++++%7C--+m.getOrElse(scala.Symbol("c"),+0)+:+pt=Double+BYVALmode-EXPRmode+(site:+value+d++in+$iw)+
%7C++++%7C++++%7C--+m.getOrElse+BYVALmode-EXPRmode-FUNmode-POLYmode+(silent:+value+d++in+$iw)+
%7C++++%7C++++%7C++++%7C--+m+EXPRmode-POLYmode-QUALmode+(silent:+value+d++in+$iw)+
%7C++++%7C++++%7C++++%7C++++\->+m.type+(with+underlying+type+scala.collection.immutable.Map[Symbol,Int])
%7C++++%7C++++%7C++++[adapt]+[B1+>:+B](key:+A,+default:+=>+B1)B1+adapted+to+[B1+>:+B](key:+A,+default:+=>+B1)B1
%7C++++%7C++++%7C++++\->+(key:+Symbol,+default:+=>+B1)B1
%7C++++%7C++++%7C--+scala.Symbol("c")+:+pt=Symbol+BYVALmode-EXPRmode-POLYmode+(site:+value+d++in+$iw)+
%7C++++%7C++++%7C++++%7C--+scala.Symbol+BYVALmode-EXPRmode-FUNmode-POLYmode+(silent:+value+d++in+$iw)+
%7C++++%7C++++%7C++++%7C++++%7C--+scala.Symbol.apply+BYVALmode-EXPRmode-FUNmode-POLYmode+(silent:+value+d++in+$iw)+
%7C++++%7C++++%7C++++%7C++++%7C++++\->+(name:+String)Symbol
%7C++++%7C++++%7C++++%7C++++[adapt]+Symbol.type+adapted+to+(name:+String)Symbol
%7C++++%7C++++%7C++++%7C++++\->+(name:+String)Symbol
%7C++++%7C++++%7C++++%7C--+"c"+:+pt=String+BYVALmode-EXPRmode+(silent:+value+d++in+$iw)+
%7C++++%7C++++%7C++++%7C++++\->+String("c")
%7C++++%7C++++%7C++++\->+Symbol
%7C++++%7C++++%7C--+0+:+pt=Double+EXPRmode-POLYmode+(site:+value+d++in+$iw)+
%7C++++%7C++++%7C++++\->+Double(0.0)
%7C++++%7C++++solving+for+(B1:+?B1)+
<console>:8:+error:+type+mismatch;
+found+++:+AnyVal
+required:+Double
+++++++val+d:+Double+=+m.getOrElse('c,+0)
++++++++++++++++++++++++++++++++++%5E
%7C++++%7C++++\->+<error>|code-block|syntax|javascript|495422|另外，-xprint:typer显示|495423|m.getOrElse[AnyVal](scala.Symbol.apply("c"),+0.0);|495424|规范解释+(case+3方法)：args是用预期的param类型键入的，这就是为什么您在那里得到到Double的转换，而不是让该方法返回一个在赋值中被扩展到Double的Int。|495425|entityMap|0|LINK|mutability|MUTABLE|url|http://www.scala-lang.org/files/archive/spec/2.11/06-expressions.html#local-type-inference^0|A|D|0|7|6|E|1|K|6|R|3|0|4|2|B|2|H|6|S|1|U|3|Y|6|0|0|3|D|0|0|1D|6|26|6|2D|3|0|4|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|1R|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1S|8|@$9|1T|A|1U|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1V|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1W|8|@$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]|$9|21|A|22|B|C]]|D|@$9|23|A|24|1|25]]|E|$]]|$1|U|3|-4|5|6|7|26|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

I'm trying to get better at reading <code>-Ytyper-debug</code> output.

With the expected type <code>Double</code>, the type of <code>0</code> is taken as <code>Double</code>, <code>0.0</code>. That's because of the widening conversions people hate.

This is before solving for <code>B1</code>. <code>B1</code> is taken as <code>AnyVal</code> because that is the lub of <code>A</code>/<code>Int</code> and <code>Double</code>.

<pre><code>| |-- Double TYPEmode (site: value d in $iw) 
| | \-&gt; Double
| |-- m.getOrElse(scala.Symbol("c"), 0) : pt=Double BYVALmode-EXPRmode (site: value d in $iw) 
| | |-- m.getOrElse BYVALmode-EXPRmode-FUNmode-POLYmode (silent: value d in $iw) 
| | | |-- m EXPRmode-POLYmode-QUALmode (silent: value d in $iw) 
| | | | \-&gt; m.type (with underlying type scala.collection.immutable.Map[Symbol,Int])
| | | [adapt] [B1 &gt;: B](key: A, default: =&gt; B1)B1 adapted to [B1 &gt;: B](key: A, default: =&gt; B1)B1
| | | \-&gt; (key: Symbol, default: =&gt; B1)B1
| | |-- scala.Symbol("c") : pt=Symbol BYVALmode-EXPRmode-POLYmode (site: value d in $iw) 
| | | |-- scala.Symbol BYVALmode-EXPRmode-FUNmode-POLYmode (silent: value d in $iw) 
| | | | |-- scala.Symbol.apply BYVALmode-EXPRmode-FUNmode-POLYmode (silent: value d in $iw) 
| | | | | \-&gt; (name: String)Symbol
| | | | [adapt] Symbol.type adapted to (name: String)Symbol
| | | | \-&gt; (name: String)Symbol
| | | |-- "c" : pt=String BYVALmode-EXPRmode (silent: value d in $iw) 
| | | | \-&gt; String("c")
| | | \-&gt; Symbol
| | |-- 0 : pt=Double EXPRmode-POLYmode (site: value d in $iw) 
| | | \-&gt; Double(0.0)
| | solving for (B1: ?B1) 
&lt;console&gt;:8: error: type mismatch;
 found : AnyVal
 required: Double
 val d: Double = m.getOrElse('c, 0)
 ^
| | \-&gt; &lt;error&gt;
</code></pre>

Also, <code>-xprint:typer</code> shows

<pre><code>m.getOrElse[AnyVal](scala.Symbol.apply("c"), 0.0);
</code></pre>

The <a href="http://www.scala-lang.org/files/archive/spec/2.11/06-expressions.html#local-type-inference" rel="nofollow">spec explains</a> (case 3 methods) that the args are typechecked with the expected type of the param, which is why you get the conversion to <code>Double</code> there, instead of having the method return an <code>Int</code> that is widened to <code>Double</code> in the assignment.

Consider

<pre><code>scala&gt; val m = Map('a -&gt; 3, 'b -&gt; 4)
m: scala.collection.immutable.Map[Symbol,Int] = Map('a -&gt; 3, 'b -&gt; 4)

scala&gt; val d: Double = m.getOrElse('c, 0)
&lt;console&gt;:8: error: type mismatch;
 found : AnyVal
 required: Double
 val d: Double = m.getOrElse('c, 0)
 ^

scala&gt; m.getOrElse('c, 0)
res0: Int = 0

scala&gt; m.getOrElse('a, 0)
res1: Int = 3
</code></pre>

Why is it that Scala thinks that the <code>getOrElse</code> call returns <code>AnyVal</code> even tho it obviously returns an <code>Int</code>?

Furthermore, even this fails with the same error:

<pre><code>scala&gt; val x: Double = m.getOrElse('a, 0): Double
&lt;console&gt;:8: error: type mismatch;
 found : AnyVal
 required: Double
 val x: Double = m.getOrElse('a, 0): Double
</code></pre>

This however works:

<pre><code>scala&gt; val x: Double = m.getOrElse('a, 0): Int
x: Double = 3.0
</code></pre>

This happens on 2.11.x; I have no tried it on 2.10.x.

Type weirdness with Map#getOrElse

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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考虑一下scala> val m = Map('a -> 3, 'b -> 4)m: scala.collection.immutable.Map[Symbol,Int] = Map('a -> 3, 'b -> 4)scala> val d: Double = m.getOrElse('c, 0)<console>:...

问Map#getOrElse的类型怪异
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问Map#getOrElse的类型怪异EN