如何以最佳方式传递元组参数? [英] How to pass a tuple argument the best way?

问题描述

如何以最佳方式传递元组参数?

How to pass a tuple argument the best way ?

示例:

def foo(...): (Int, Int) = ...

def bar(a: Int, b: Int) = ...

现在我想将 foo 的输出传递给 bar.这可以通过以下方式实现:

Now I would like to pass the output of foo to bar. This can be achieved with:

val fooResult = foo(...)
bar(fooResult._1, fooResult._2)

这种方法看起来有点难看，尤其是当我们用 n > 处理 n-元组时.2.此外，我们必须将 foo 的结果存储在一个额外的值中，否则必须使用 bar(foo._1, foo._2) 多次执行 foo.

This approach looks a bit ugly, especially when we deal with a n-tuple with n > 2. Also we have to store the result of foo in an extra value, because otherwise foo has to be executed more than once using bar(foo._1, foo._2).

是否有更好的方法将元组作为参数传递?

推荐答案

每个函数都有一个特殊的tupled方法:

There is a special tupled method available for every function:

val bar2 = (bar _).tupled  // or Function.tupled(bar _)

bar2 采用 (Int, Int) 元组(与 bar 参数相同).现在你可以说:

bar2 takes a tuple of (Int, Int) (same as bar arguments). Now you can say:

bar2(foo())

如果你的方法实际上是函数(注意 val 关键字)，语法会更令人愉快:

If your methods were actually functions (notice the val keyword) the syntax is much more pleasant:

val bar = (a: Int, b: Int) => //...
bar.tupled(foo())

另见

• 如何将函数应用于元组?

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