blocks|key|1479690|text|您可以使用Java库：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1479691|val+number+=+Integer.parseInt("FFFF",+16)
>+number:+Int+=+65535|code-block|syntax|javascript|1479692|或者如果你感到兴奋:-)：|1479693|implicit+def+hex2int+(hex:+String):+Int+=+Integer.parseInt(hex,+16)

val+number:+Int+=+"CAFE"+//+<-+behold+the+magic
number:+Int+=+51966|1479694|1479695|此外，如果您不是专门尝试将字符串参数解析为十六进制，请注意Scala直接支持十六进制整数字面量。在这种情况下：|1479696|val+x+=+0xCAFE
>+x:+Int+=+51966|1479697|Scala是不是很棒？:-)|1479698|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|Y|8|@]|9|@]|A|$]]|$1|N|3|O|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|P|3|Q|5|6|7|10|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|S|$]]

You can use the Java libs:

<pre><code>val number = Integer.parseInt("FFFF", 16)
&gt; number: Int = 65535
</code></pre>

Or if you are feeling sparky :-):

<pre><code>implicit def hex2int (hex: String): Int = Integer.parseInt(hex, 16)

val number: Int = "CAFE" // &lt;- behold the magic
number: Int = 51966
</code></pre>

<hr>

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:

<pre><code>val x = 0xCAFE
&gt; x: Int = 51966
</code></pre>

Isn't Scala wonderful? :-)

blocks|key|2576412|text|7zark7的答案是正确的，但我想做一些补充。从String隐含到Int可能是危险的。相反，您可以对包装器使用隐式转换，并显式调用解析：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2576413|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|code-block|syntax|javascript|2576414|entityMap^0|O|6|X|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

7zark7 answer is correct, but I want to make some additions.
Implicit from <code>String</code> to <code>Int</code> can be dangerous. Instead you can use implicit conversion to wrapper and call parsing explicitly:

<pre><code>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
</code></pre>

blocks|key|2576433|text|那么一行代码呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2576434|def+hexToInt(s:+String):+Int+=+{
++++s.toList.map("0123456789abcdef".indexOf(_)).reduceLeft(_+*+16+%2B+_)
}

scala>+hexToInt("cafe")
res0:+Int+=+51966|code-block|syntax|javascript|2576435|回答你的第二个问题：|2576436|2576437|是不是有类似于“A”.toInt(基础)的东西？|blockquote|2576438|2576439|2576440|是的，仍然是一行：|2576441|def+baseToInt(s:+String,+base:+String):+Int+=+{
++++s.toList.map(base.indexOf(_)).reduceLeft(_+*+base.length+%2B+_)
}

scala>+baseToInt("1100",+"01")
res1:+Int+=+12|2576442|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|Y|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|11|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|T|$]]

What about a one-liner?

<pre><code>def hexToInt(s: String): Int = {
 s.toList.map("0123456789abcdef".indexOf(_)).reduceLeft(_ * 16 + _)
}

scala&gt; hexToInt("cafe")
res0: Int = 51966
</code></pre>

And to answer your second item: 

<blockquote>
 Is there something like "A".toInt(base)?
</blockquote>

Yes, still as a one-liner:

<pre><code>def baseToInt(s: String, base: String): Int = {
 s.toList.map(base.indexOf(_)).reduceLeft(_ * base.length + _)
}

scala&gt; baseToInt("1100", "01")
res1: Int = 12
</code></pre>

blocks|key|1480136|text|任何想要将UUID从十六进制转换为十进制数的人都可以从Benoit的答案中借用，并使用BigDecimal来完成这项工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1480137|scala>+"03cedf84011dd11e38ff0800200c9a66".toList.map(
+%7C+++"0123456789abcdef".indexOf(_)).map(
+%7C+++++BigInt(_)).reduceLeft(+_+*+16+%2B+_)
res0:+scala.math.BigInt+=+5061830576017519706280227473241971302|code-block|syntax|javascript|1480138|或者更一般地说：|1480139|def+hex2dec(hex:+String):+BigInt+=+{
++hex.toLowerCase().toList.map(
++++"0123456789abcdef".indexOf(_)).map(
++++BigInt(_)).reduceLeft(+_+*+16+%2B+_)
}

def+uuid2dec(uuid:+UUID):+BigInt+=+{
++hex2dec(uuid.toString.replace("-",""))
}|1480140|然后：|1480141|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|1480142|这样做的一个实际应用是在条形码中编码UUID，其中Code128为所有数字生成的条形码比字母数字字符串生成的条形码更短。请参阅http://en.wikipedia.org/wiki/Code128#Subtypes上有关"128A“子类型的说明。|offset|length|1480143|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Code128#Subtypes^0|0|0|0|0|0|0|1R|19|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@$Q|17|R|18|1|19]]|A|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Anyone wanting to convert a UUID from hex to a decimal number can borrow from Benoit's answer and use BigDecimal for the job:

<pre><code>scala&gt; "03cedf84011dd11e38ff0800200c9a66".toList.map(
 | "0123456789abcdef".indexOf(_)).map(
 | BigInt(_)).reduceLeft( _ * 16 + _)
res0: scala.math.BigInt = 5061830576017519706280227473241971302
</code></pre>

Or more generally:

<pre><code>def hex2dec(hex: String): BigInt = {
 hex.toLowerCase().toList.map(
 "0123456789abcdef".indexOf(_)).map(
 BigInt(_)).reduceLeft( _ * 16 + _)
}

def uuid2dec(uuid: UUID): BigInt = {
 hex2dec(uuid.toString.replace("-",""))
}
</code></pre>

Then:

<pre><code>scala&gt; import java.util.UUID

scala&gt; val id = UUID.fromString("3CEDF84-011D-D11E-38FF-D0800200C9A66")
id: java.util.UUID = 03cedf84-011d-d11e-38ff-0800200c9a66

scala&gt; uuid2dec(id)
res2: BigInt = 5061830576017519706280227473241971302
</code></pre>

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 <a href="http://en.wikipedia.org/wiki/Code128#Subtypes" rel="noreferrer">http://en.wikipedia.org/wiki/Code128#Subtypes</a>.

blocks|key|2492487|text|无论长短，也可以直接使用Java方法，比如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2492488|Long2long(java.lang.Long.valueOf(hexString,+16))|code-block|syntax|javascript|2492489|在某些情况下甚至可以省略Long2long。|offset|length|style|CODE|2492490|entityMap^0|0|0|C|9|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

For Long and Short, it is also possible to use the Java methods directly like

<pre><code>Long2long(java.lang.Long.valueOf(hexString, 16))
</code></pre>

where <code>Long2long</code> can be even be omitted in some cases.

Just can't find a way to transform an Hex String to a number (Int, Long, Short) in Scala.

Is there something like <code>"A".toInt(base)</code>?

Hex String to Int,Short and Long in Scala

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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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。有没有像"A".toInt(base)这样的东西

问在Scala中将十六进制字符串转换为Int、Short和Long
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回答 5

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