字符或字符串 ->Scala 中的 Unicode 值? [英] Char or String -> Unicode value in Scala?

问题描述

所以我正在完成不耐烦的 Scala"中的一些练习，其中之一是:

So I'm working through a few of the exercises in "Scala for the Impatient" and one of them is:

编写一个 for 循环来计算字符串中所有字母的 Unicode 代码的乘积.例如，你好"中的字符的乘积为 9415087488 L.

Write a for loop for computing the product of the Unicode codes of all letters in a string. For example, the product of the characters in "Hello" is 9415087488 L.

下一个问题是做同样的事情，但没有 for 循环 - 它暗示我们应该检查 Scaladoc 中的 StringOps.

The next problem is to do the same, but without a for loop - it hints that we should check StringOps in Scaladoc.

我检查了 Scaladoc 中的 RichChar 和 StringOps 部分，也许我读错了或找错了地方，但我找不到任何让我心动的东西以匹配他们的输出.到目前为止，我已经尝试过:

I checked the RichChar and StringOps section in Scaladoc, and perhaps I'm misreading or looking in the wrong places, but I can't find anything that gets me to match their output. I've thus far tried:

scala> x.foldLeft(1)(_ * _.toInt)
res0: Int = 825152896

scala> x.foldLeft(1)(_ * _.getNumericValue)
res5: Int = 2518992

scala> x.foldLeft(1)(_ * _.intValue())
res6: Int = 825152896

scala> var x = 1
x: Int = 1

scala> for (c <- "Hello") x *= c.toInt

scala> x
res12: Int = 825152896

与他们的输出不匹配.

如何以 for 和非 for 方式执行此操作?

How do I do this, in both the for and non-for way?

谢谢！

推荐答案

当你做 x.foldLeft(1)(_ * _.toInt) 时，结果类型将推断为 x.foldLeft(1)(_ * _.toInt)code>Int，但 9415087488 太大，Int 无法存储它.

When you do x.foldLeft(1)(_ * _.toInt), the result type will be inference to an Int, but 9415087488 is too large for an Int to store it.

所以你需要告诉 Scala 使用 Long 来存储它.

So you need to tell Scala using Long to store it.

scala> val x = "Hello"
x: java.lang.String = Hello

scala> x.foldLeft(1L)(_ * _.toInt)
res1: Long = 9415087488

scala> var x: Long = 1
x: Long = 1

scala> for (c <- "Hello") x *= c.toInt

scala> x
res7: Long = 9415087488

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