Java 加减字符为什么会这样？（爪哇）

adding and subtraction chars why does this work? (java)

提问by thrax

I'm learning Java through "introduction to Java programming 9th edition" by Daniel liang at chapter 9 "strings" I've encountered this piece of code :

我正在通过 Daniel liang 在第 9 章“字符串”的“Java 编程简介第 9 版”学习 Java，我遇到了这段代码：

    public static int hexCharToDecimal(char ch){
       if (ch >= 'A' && ch <= 'F')
           return 10 + ch - 'A';
       else
          return ch - '0';
}

oki so can someone explain what just happened in here? how is possible to add,sub chars from integers and what's the meaning behind it and thanks.

oki 所以有人可以解释一下这里发生了什么吗？如何从整数中添加子字符以及它背后的含义是什么，谢谢。

采纳答案by Michael Yaworski

From the Docs

来自文档

The char data type is a single 16-bit Unicode character.

char 数据类型是单个 16 位 Unicode 字符。

A charis represented by its code point value:

Achar由其代码点值表示：

• min '\u0000'(or 0)
• max: '\uffff'(or 65,535)

• 最小'\u0000'（或 0）
• 最大值：（'\uffff'或 65,535）

You can see all of the English alphabetic code points on an ASCII table.

您可以在ASCII 表上看到所有英文字母代码点。

_{Note that 0 == \u0000and 65,535 == \uffff, as well as everything in between. They are corresponding values.}

_{请注意0 == \u0000和65,535 == \uffff，以及介于两者之间的所有内容。它们是对应的值。}

A charis actually just stored as a number (its code point value). We have syntax to represent characters like char c = 'A';, but it's equivalent to char c = 65;and 'A' == 65is true.

Achar实际上只是存储为一个数字（它的代码点值）。我们有语法来表示像 那样的字符char c = 'A';，但它等价于char c = 65;并且'A' == 65为真。

So in your code, the chars are being represented by their decimal values to do arithmetic (whole numbers from 0 to 65,535).

因此，在您的代码中，字符由它们的十进制值表示以进行算术运算（从 0 到 65,535 的整数）。

For example, the char 'A'is represented by its code point 65(decimal value in ASCII table):

例如，char'A'由其代码点65（ASCII 表中的十进制值）表示：

System.out.print('A'); // prints A
System.out.print((int)('A')); // prints 65 because you casted it to an int

_{As a note, a shortis a 16-bit signedinteger, so even though a charis also 16-bits, the maximum integer value of a char(65,535) exceeds the maximum integer value of a short(32,767). Therefore, a cast to (short)from a charcannot always work. And the minimum integer value of a charis 0, whereas the minimum integer value of a shortis -32,768.}

_{注意，ashort是一个 16 位有符号整数，所以即使 achar也是 16 位，a 的最大整数值char(65,535) 超过了 a short(32,767)的最大整数值。因此，(short)从 a的强制转换char并不总是有效。并且a的最小整数值为char0，而a的最小整数值为short-32,768。}

---

---

For your code, let's say that the charwas 'D'. Note that 'D' == 68since its code point is 68.

对于您的代码，假设charwas 'D'。请注意，'D' == 68由于它的代码点是68.

return 10 + ch - 'A';

This returns 10 + 68 - 65, so it will return 13.

这将返回10 + 68 - 65，因此它将返回13。

Now let's say the char was 'Q' == 81.

现在让我们说 char 是'Q' == 81.

if (ch >= 'A' && ch <= 'F')

This is false since 'Q' > 'F'(81 > 70), so it would go into the elseblock and execute:

这是错误的，因为'Q' > 'F'( 81 > 70)，所以它会进入else块并执行：

return ch - '0';

This returns 81 - 48so it will return 33.

这将返回，81 - 48因此它将返回33。

Your function returns an inttype, but if it were to instead return a charor have the intcasted to a charafterward, then the value 33returned would represent the '!'character, since 33is its code point value. Look up the character in ASCII table or Unicode table to verify that '!' == 33(compare decimal values).

您的函数返回一个int类型，但如果它改为返回 achar或在之后将其int强制转换为 a char，则33返回的值将代表该'!'字符，因为33它是其代码点值。在 ASCII 表或 Unicode 表中查找字符以验证'!' == 33（比较十进制值）。

回答by praveen_mohan

Chars are in turn stored as integers(ASCII value) so that you can perform add and sub on integers which will return ASCII value of a char

字符依次存储为整数（ASCII 值），以便您可以对整数执行加法和减法，这将返回字符的 ASCII 值

回答by ejbs

This is because char is a primitive type which can be used as a numerical value. Every character in a string is encoded as a specific number (not entirely true in all cases, but good enough for a basic understanding of the matter) and Java allows you to use chars in such a way.

这是因为 char 是一种原始类型，可以用作数值。字符串中的每个字符都被编码为一个特定的数字（并非在所有情况下都完全正确，但足以对问题有基本的了解）并且 Java 允许您以这种方式使用字符。

It probably allows this mostly for historical reasons, this is how it worked in C and they probably motivated it with "performance" or something like that.

它可能主要出于历史原因允许这样做，这就是它在 C 中的工作方式，并且他们可能以“性能”或类似的东西来激励它。

If you think it's weird then don't worry, I think so too

如果你觉得这很奇怪那么别担心，我也这么认为

The other answer is incorrect actually. ASCII is a specific encoding (an encoding is some specification that says "1 = A, 2 = B, ... , 255 = Space") and that is not the one used in Java. A Java char is two bytes wide and is interpreted through the unicode character encoding.

另一个答案实际上是不正确的。ASCII 是一种特定的编码（一种编码是一些规范，上面写着“1 = A, 2 = B, ... , 255 = Space”），而不是 Java 中使用的编码。Java char 是两个字节宽，通过 unicode 字符编码进行解释。

回答by nhgrif

Regardless of how Java actually stores the chardatatype, what's certain is this, the character 'A'subtracted from the character 'A'would be represented as the nullcharacter, \0. In memory, this means every bit is 0. The size in memory a chartakes up in memory may vary from language to language, but as far as I know, the nullcharacter is the same in all the languages, every bit is equal to 0.

不管 Java 实际如何存储char数据类型，可以肯定的是，从字符中'A'减去的字符'A'将表示为null字符\0. 在内存中，这意味着每一位都是0. 内存中 a 在内存中char所占的大小可能因语言而异，但据我所知，null所有语言中的字符都是相同的，每一位都等于0.

As an intvalue, a piece of memory with every bit equal to 0represents the integer value of 0.

作为一个int值，一块每一位都等于的内存0代表0的整数值。

And as it turns out, when you do "character math", subtracting any alphabetical character from any other alphabetical character (of the same case) results in bits being flipped in such a way that, if you were to interpret them as an int, would represent the distance between these characters. Additionally, subtracting the char '0'from any other numeric char will result in int value of the char you subtracted from, for basically the same reason.

事实证明，当您进行“字符数学”时，从任何其他字母字符（相同情况下）中减去任何字母字符会导致位以这样的方式翻转，如果您将它们解释为int，代表这些字符之间的距离。此外，'0'从任何其他数字 char 中减去char 将导致您从中减去的 char 的 int 值，原因基本相同。

'A' - 'A' = '##代码##'
'a' - 'a' = '##代码##'
'0' - '0' = '##代码##'