为什么我得到一个“缺少扩展函数的参数"?在一种情况下而不是另一种情况? [英] Why do I get a "missing parameter for expanded function" in one case and not the other?

问题描述

这种情况下:

Seq(fromDir, toDir) find (!_.isDirectory) foreach (println(_))

然而这不是:

Seq(fromDir, toDir) find (!_.isDirectory) foreach (throw new Exception(_.toString))

编译以这个错误结束:

error: missing parameter type for expanded function ((x$4) => x$4.toString)

现在如果我这样写，它会再次编译:

Now if I write it this way it compiles again:

Seq(fromDir, toDir) find (!_.isDirectory) foreach (s => throw new Exception(s.toString))

我相信有一个合理的解释;)

I am sure there is a reasonable explanation ;)

推荐答案

这已经在 一个相关问题.下划线向外延伸到最接近的结束 Expr :顶级表达式或括号中的表达式.

This has already been addressed in a related question. Underscores extend outwards to the closest closing Expr : top-level expressions or expressions in parentheses.

(_.toString) 是括号中的表达式.因此，您在错误情况下传递给 Exception 的参数在扩展后是完整的匿名函数 (x$1) =>A <: Any => 类型的 x$1.toStringString，而 Exception 需要一个 String.

(_.toString) is an expression in parentheses. The argument you are passing to Exception in the error case is therefore, after expansion, the full anonymous function (x$1) => x$1.toString of type A <: Any => String, while Exception expects a String.

在 println 的情况下，_ 本身不属于语法类别 Expr，而是 (println (_)) 是，所以你得到预期的 (x$0) =>println(x$0).

In the println case, _ by itself isn't of syntactic category Expr, but (println (_)) is, so you get the expected (x$0) => println(x$0).

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