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Home/ Questions/Q 3931938
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T23:24:06+00:00 2026-05-19T23:24:06+00:00

Just wanted to know question w.r.t Currying If we have defined the curried function

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Just wanted to know question w.r.t Currying

If we have defined the curried function curriedNewSum

scala> def curriedNewSum(x : Int)(y : Int) = x + y
curriedNewSum: (x: Int)(y: Int)Int

scala> curriedNewSum(10)(20)
res5: Int = 30

scala> var tenPlus = curriedNewSum(10)_
tenPlus: (Int) => Int = <function1>

scala> tenPlus(20)
res6: Int = 30

scala> var plusTen = curriedNewSum(_)(20)
<console>:6: error: missing parameter type for expanded function ((x$1) => curri
edNewSum(x$1)(20))
       var plusTen = curriedNewSum(_)(20)
                                   ^

So why does curriedNewSum(10)_ works & curriedNewSum(_)(10) not?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-19T23:24:07+00:00Added an answer on May 19, 2026 at 11:24 pm

    I’m not 100% sure what exactly is the problem, but I strongly suspect this isn’t doing what you think it is.

    Try, for instance,

    var plusTen = curriedNewSum(_)
    

    You’ll see it will return a Function1[Int, Function1[Int, Int]]. Now try this:

    var plusTen = (curriedNewSum(_))(10)
    

    And see it work! Well, that translates into:

    var plusTen = ((x: Int) => curriedNewSum(x))(10)
    

    While the other way translates into:

    var plusTen = (x) => curriedNewSum(x)(10)
    

    Something about how the function is expanding is screwing up with the type inference.

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