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    <p>If you are thinking about partial application, I thought that this was only possible with <strong>multiple parameter lists</strong> (whereas you only have one):</p> <pre><code>def plus(x: Int)(y: Int) = x + y //x and y in different parameter lists val plus10 = plus(10) _ //_ indicates partial application println(plus10(2)) //prints 12 </code></pre> <p>Your example is interesting though as I was completely unaware of the syntax you describe and it appears you can have partial application with a single parameter list:</p> <pre><code>scala&gt; def plus2(x: Int, y: Int) = x + y plus2: (x: Int,y: Int)Int scala&gt; val anon = plus2(_,_) anon: (Int, Int) =&gt; Int = &lt;function2&gt; scala&gt; anon(3, 4) res1: Int = 7 </code></pre> <p>So the compiler can clearly infer the type <code>Int</code>!</p> <pre><code>scala&gt; val anon2 = plus2(20,_) &lt;console&gt;:5: error: missing parameter type for expanded function ((x$1) =&gt; plus2(20, x$1)) val anon2 = plus2(20,_) ^ </code></pre> <p>Hmmm, strange! I don't seem to be able to do partial application with a single parameter list. But then if I declare the type of the second parameter, I <em>can</em> have partial application!</p> <pre><code>scala&gt; val anon2 = plus2(20,_: Int) anon2: (Int) =&gt; Int = &lt;function1&gt; scala&gt; anon2(24) res2: Int = 44 </code></pre> <p><strong>EDIT</strong> - one thing I would observe is that it seems like the following two shortenings are equivalent, in which case it's a bit more obvious that this is not a "partial application" but more like a "function pointer"</p> <pre><code>val anon1 = plus2(_,_) val anon2 = plus2 _ </code></pre>
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    1. POIn Scala, why can't I partially apply a function without explicitly specifying its argument types?
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    1. CO`val f = foo(_, _)` expands to `val f = (a, b) => foo(a, b)`, it isn't precisely the same as partially applying a curried function.
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    2. COBut it *seems* to have the same effect, you must admit. And if the compiler can infer types with 2 `_` placeholders, why does it have trouble when there is only one?
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    3. COTo answer that question, you could debug through Infer.scala in the compiler :) I agree, it seems like an unnecessary limitation, but scala's type inference is intentionally omitted from the spec to allow such wiggle room.
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