在 Scala 中将十六进制字符串转换为 Int、Short 和 Long
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Hex String to Int,Short and Long in Scala
提问by rsan
Just can't find a way to transform an Hex String to a number (Int, Long, Short) in Scala.
只是找不到在 Scala 中将十六进制字符串转换为数字（Int、Long、Short）的方法。
Is there something like "A".toInt(base)?
有类似的东西"A".toInt(base)吗？
回答by 7zark7
You can use the Java libs:
您可以使用 Java 库：
val number = Integer.parseInt("FFFF", 16)
> number: Int = 65535
Or if you are feeling sparky :-):
或者，如果你感觉闪闪发光:-)：
implicit def hex2int (hex: String): Int = Integer.parseInt(hex, 16)
val number: Int = "CAFE" // <- behold the magic
number: Int = 51966
Also, if you aren't specifically trying to parse a String parameter into hex, note that Scala directly supports hexadecimal Integer literals. In this case:
此外，如果您不是专门尝试将 String 参数解析为十六进制，请注意 Scala 直接支持十六进制整数文字。在这种情况下：
val x = 0xCAFE
> x: Int = 51966
Isn't Scala wonderful? :-)
Scala 是不是很棒？:-)
回答by Sergey Passichenko
7zark7 answer is correct, but I want to make some additions.
Implicit from Stringto Intcan be dangerous. Instead you can use implicit conversion to wrapper and call parsing explicitly:
7zark7 答案是正确的，但我想补充一些。从隐String到Int可能是危险的。相反，您可以使用隐式转换来包装并显式调用解析：
class HexString(val s: String) {
def hex = Integer.parseInt(s, 16)
}
implicit def str2hex(str: String): HexString = new HexString(str)
val num: Int = "CAFE".hex
回答by Benoit
What about a one-liner?
单线呢？
def hexToInt(s: String): Int = {
s.toList.map("0123456789abcdef".indexOf(_)).reduceLeft(_ * 16 + _)
}
scala> hexToInt("cafe")
res0: Int = 51966
And to answer your second item:
并回答你的第二个项目：
Is there something like "A".toInt(base)?
有没有像 "A".toInt(base) 这样的东西？
Yes, still as a one-liner:
是的，仍然作为单线：
def baseToInt(s: String, base: String): Int = {
s.toList.map(base.indexOf(_)).reduceLeft(_ * base.length + _)
}
scala> baseToInt("1100", "01")
res1: Int = 12
回答by Andrew E
Anyone wanting to convert a UUID from hex to a decimal number can borrow from Benoit's answer and use BigDecimal for the job:
任何想要将 UUID 从十六进制转换为十进制数的人都可以借鉴 Benoit 的答案并使用 BigDecimal 来完成这项工作：
scala> "03cedf84011dd11e38ff0800200c9a66".toList.map(
| "0123456789abcdef".indexOf(_)).map(
| BigInt(_)).reduceLeft( _ * 16 + _)
res0: scala.math.BigInt = 5061830576017519706280227473241971302
Or more generally:
或更一般地：
def hex2dec(hex: String): BigInt = {
hex.toLowerCase().toList.map(
"0123456789abcdef".indexOf(_)).map(
BigInt(_)).reduceLeft( _ * 16 + _)
}
def uuid2dec(uuid: UUID): BigInt = {
hex2dec(uuid.toString.replace("-",""))
}
Then:
然后：
scala> import java.util.UUID
scala> val id = UUID.fromString("3CEDF84-011D-D11E-38FF-D0800200C9A66")
id: java.util.UUID = 03cedf84-011d-d11e-38ff-0800200c9a66
scala> uuid2dec(id)
res2: BigInt = 5061830576017519706280227473241971302
One practical application for this is encoding the UUID in a barcode, where Code128 produces a shorter barcode for all digits than it does with alphanumeric strings. See notes about subtype "128A" on http://en.wikipedia.org/wiki/Code128#Subtypes.
对此的一个实际应用是在条形码中编码 UUID，其中 Code128 为所有数字生成一个比使用字母数字字符串更短的条形码。请参阅http://en.wikipedia.org/wiki/Code128#Subtypes上有关子类型“128A”的注释。
回答by oseiskar
For Long and Short, it is also possible to use the Java methods directly like
对于 Long 和 Short，也可以直接使用 Java 方法，如
Long2long(java.lang.Long.valueOf(hexString, 16))
where Long2longcan be even be omitted in some cases.
Long2long在某些情况下甚至可以省略where 。
