blocks|key|537061|text|对于int+(带符号的4字节)来说，它太大了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|537062|使用|537063|Long.parseLong("AA0F245C",+16);|code-block|syntax|javascript|537064|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

It's simply too big for an int (which is 4 bytes and signed).

Use 

<pre><code>Long.parseLong("AA0F245C", 16);
</code></pre>

blocks|key|320224|text|一个Java+Integer可以处理的最大值是2147483657，即2%5E31-1。十六进制数AA0F245C的十进制数是2853119068，太大了，因此需要使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|320225|Long.parseLong("AA0F245C",+16);|code-block|syntax|javascript|320226|才能让它工作。|320227|entityMap^0|7|7|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

The maximum value that a Java <code>Integer</code> can handle is 2147483657, or 2^31-1. The hexadecimal number AA0F245C is 2853119068 as a decimal number, and is far too large, so you need to use 

<pre><code>Long.parseLong("AA0F245C", 16);
</code></pre>

to make it work.

blocks|key|320295|text|你可以通过带有格式参数的parseInt很容易地做到这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|320296|Integer.parseInt("-FF",+16)+;+//+returns+-255|code-block|syntax|javascript|320297|javadoc+Integer|offset|length|320298|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#parseInt(java.lang.String,%2520int)^0|0|0|0|F|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

you can easily do it with parseInt with format parameter.
<pre><code>Integer.parseInt(&quot;-FF&quot;, 16) ; // returns -255
</code></pre>
<a href="https://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#parseInt(java.lang.String,%20int)" rel="nofollow noreferrer">javadoc Integer</a>

blocks|key|320313|text|你可以像这样使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|320314|System.out.println(Integer.decode("0x4d2"))++++//+output+1234
//and+vice+versa+
System.out.println(Integer.toHexString(1234);+//++output+is+4d2);|code-block|syntax|javascript|320315|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you may use like that 

<pre><code>System.out.println(Integer.decode("0x4d2")) // output 1234
//and vice versa 
System.out.println(Integer.toHexString(1234); // output is 4d2);
</code></pre>

blocks|key|537255|text|这是正确的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|537256|myPassedColor+=+"#ffff8c85"+int+colorInt+=+Color.parseColor(myPassedColor)|offset|length|style|CODE|537257|entityMap^0|0|0|22|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

This is the right answer: 

<code>myPassedColor = "#ffff8c85"
int colorInt = Color.parseColor(myPassedColor)</code>

blocks|key|320421|text|对于那些需要将有符号字节的十六进制表示形式从两个字符字符串转换为字节(在Java中总是有符号的)的人，有一个示例。解析十六进制字符串永远不会给出负数，这是错误的，因为从某种角度来看，0xFF是-1+(二的补码编码)。其原理是将传入的字符串解析为int，它大于byte，然后绕过负数。我只显示了字节，所以这个例子足够简短。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|320422|String+inputTwoCharHex="FE";+//whatever+your+microcontroller+data+is

int+i=Integer.parseInt(inputTwoCharHex,16);
//negative+numbers+is+i+now+look+like+128...255

//+shortly+i>=128
if+(i>=Integer.parseInt("80",16)){

++++//need+to+wrap-around+negative+numbers
++++//we+know+that+FF+is+255+which+is+-1
++++//and+FE+is+254+which+is+-2+and+so+on
++++i=-1-Integer.parseInt("FF",16)%2Bi;
++++//shortly+i=-256%2Bi;
}
byte+b=(byte)i;
//b+is+now+surely+between+-128+and+%2B127|code-block|syntax|javascript|320423|可以对其进行编辑，以处理更长的数字。只需分别添加更多FF或00即可。对于解析8个十六进制字符的有符号整数，您需要使用Long.parseLong，因为FFFF-FFFF是整数-1，在表示为正数时不适合integer+(给出4294967295)。所以你需要很长的时间来存储它。在转换为负数并转换回Integer后，它将适合。没有8个字符的十六进制字符串，最后不适合整数。|320424|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

For those of you who need to convert hexadecimal representation of a signed byte from two-character String into byte (which in Java is always signed), there is an example. Parsing a hexadecimal string never gives negative number, which is faulty, because 0xFF is -1 from some point of view (two's complement coding). The principle is to parse the incoming String as int, which is larger than byte, and then wrap around negative numbers.
I'm showing only bytes, so that example is short enough.

<pre><code>String inputTwoCharHex="FE"; //whatever your microcontroller data is

int i=Integer.parseInt(inputTwoCharHex,16);
//negative numbers is i now look like 128...255

// shortly i&gt;=128
if (i&gt;=Integer.parseInt("80",16)){

 //need to wrap-around negative numbers
 //we know that FF is 255 which is -1
 //and FE is 254 which is -2 and so on
 i=-1-Integer.parseInt("FF",16)+i;
 //shortly i=-256+i;
}
byte b=(byte)i;
//b is now surely between -128 and +127
</code></pre>

This can be edited to process longer numbers. Just add more FF's or 00's respectively. For parsing 8 hex-character signed integers, you need to use Long.parseLong, because FFFF-FFFF, which is integer -1, wouldn't fit into Integer when represented as a positive number (gives 4294967295). So you need Long to store it. After conversion to negative number and casting back to Integer, it will fit. There is no 8 character hex string, that wouldn't fit integer in the end.

blocks|key|537317|text|//Method+for+Smaller+Number+Range:
Integer.parseInt("abc",16);

//Method+for+Bigger+Number+Range.
Long.parseLong("abc",16);

//Method+for+Biggest+Number+Range.
new+BigInteger("abc",16);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|537318|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>//Method for Smaller Number Range:
Integer.parseInt("abc",16);

//Method for Bigger Number Range.
Long.parseLong("abc",16);

//Method for Biggest Number Range.
new BigInteger("abc",16);
</code></pre>

blocks|key|320482|text|除了所建议的选项之外，还有一个选项是使用BigInteger类。因为十六进制值通常是很大的数字，比如sha256或sha512值，它们很容易溢出int和long。正如其他答案所示，虽然转换为字节数组是一种选择，但java中经常被遗忘的类BigInterger也是一种选择。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|320483|String+sha256+=+"65f4b384672c2776116d8d6533c69d4b19d140ddec5c191ea4d2cfad7d025ca2";
BigInteger+value+=+new+BigInteger(sha256,+16);
System.out.println("value+=+"+%2B+value);
//+46115947372890196580627454674591875001875183744738980595815219240926424751266|code-block|syntax|javascript|320484|entityMap^0|K|A|20|3|24|4|3A|B|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

An additional option to the ones suggested, is to use the <code>BigInteger</code> class. Since hex values are often large numbers, such as sha256 or sha512 values, they will easily overflow an <code>int</code> and a <code>long</code>. While converting to a byte array is an option as other answers show, <code>BigInterger</code>, the often forgotten class in java, is an option as well.

<pre><code>String sha256 = "65f4b384672c2776116d8d6533c69d4b19d140ddec5c191ea4d2cfad7d025ca2";
BigInteger value = new BigInteger(sha256, 16);
System.out.println("value = " + value);
// 46115947372890196580627454674591875001875183744738980595815219240926424751266
</code></pre>

blocks|key|537425|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|537426|long+abc=convertString2Hex("1A2A3B")；|537427|private++long++convertString2Hex(String+numberHexString)
{
++++char[]+ChaArray+=+numberHexString.toCharArray();
++++long+HexSum=0;
++++long+cChar+=0;

++++for(int+i=0;i<numberHexString.length();i%2B%2B+)
++++{
++++++++if(+(ChaArray[i]>='0')+&&+(ChaArray[i]<='9')+)
++++++++++++cChar+=+ChaArray[i]+-+'0';
++++++++else
++++++++++++cChar+=+ChaArray[i]-'A'%2B10;
++++++++HexSum+=+16+*+HexSum+%2B+cChar;
++++}
++++return++HexSum;
}|code-block|syntax|javascript|537428|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try with this:

long abc=convertString2Hex("1A2A3B");

<pre><code>private long convertString2Hex(String numberHexString)
{
 char[] ChaArray = numberHexString.toCharArray();
 long HexSum=0;
 long cChar =0;

 for(int i=0;i&lt;numberHexString.length();i++ )
 {
 if( (ChaArray[i]&gt;='0') &amp;&amp; (ChaArray[i]&lt;='9') )
 cChar = ChaArray[i] - '0';
 else
 cChar = ChaArray[i]-'A'+10;
 HexSum = 16 * HexSum + cChar;
 }
 return HexSum;
}
</code></pre>

I am trying to convert a string that is 8 characters long of hex code into an integer so that I can do int comparison instead of string comparisons over a lot of different values.

I know this is fairly trivial in C++ but I need to do it in Java. The test case I need to satisfy is essentially to Convert "AA0F245C" to int and then back to that string so that I know it is converting right.

I have tried the following:

<pre><code>int decode = Integer.decode("0xAA0F245C"); // NumberFormatException
int decode2 = Integer.decode("AA0F245C"); //NumberFormatException
int parseInt = Integer.parseInt("AA0F245C", 16); //NumberFormatException
int valueOf = Integer.valueOf("AA0F245C", 16); //NumberFormatException
</code></pre>

I have also tried doing it two characters at a time and multiplying the results, which the conversion works but the number is not right.

<pre><code>int id = 0;
for (int h = 0; h &lt; hex.length(); h= h+2)
{
 String sub = hex.subSequence(h, h+2).toString();

if (id == 0)
 id = Integer.valueOf(sub, 16);
else
 id *= Integer.valueOf(sub, 16); 
 }
//ID = 8445600 which = 80DEA0 if I convert it back. 
</code></pre>

I can not use third party libraries just so you know, so this has to be done in Java standard libraries. 

Thank you for your help in advance.

Convert hex string to int

Java

我正在尝试将一个8个字符长度的十六进制代码的字符串转换为一个整数，这样我就可以进行int比较，而不是对许多不同的值进行字符串比较。我知道这在C++中是相当微不足道的，但我需要用Java语言来做。我需要满足的测试用例基本上是将"AA0F245C“转换为整数，然后再转换回该字符串，这样我就知道它是正确转换的。我尝试过以下几...

问将十六进制字符串转换为int
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将十六进制字符串转换为intEN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问将十六进制字符串转换为int
EN