C / C ++中的十六进制转义码中的位数 [英] number of digits in a hex escape code in C/C++

问题描述

我和我的同事发生争执。她说以下内容：

  char * a =\x000aaxz; 
  

编译器可以看到\x000aa。我不同意她的意见，因为我认为你可以在 \x 之后最多可以有4个十六进制字符。你可以有超过4个十六进制字符吗？

谁在这里？

解决方案>

§2.13.2/ 4：

逃脱\xhhh由
反斜杠后跟x跟随通过
一个或多个十六进制数字，
用于指定
所需字符的值。
十六进制序列中的数字位数没有限制
。
八进制或十六进制数字的序列分别由
的第一个字符终止，而不是八进制数字或十六进制
数字。

她是对的。

但是，您可以通过急切的期望来终止它：文字的顺序\x000aaxz指定单个四字符字符串文字。 （2.13.4 / 3）

另请注意，Unicode 使用21位代码点;它不会停止在16位。

I'm having a dispute with a colleague of mine. She says that the following:

char* a = "\x000aaxz";

will/can be seen by the compiler as "\x000aa". I do not agree with her, as I think you can have a maximum number of 4 hex characters after the \x. Can you have more than 4 hex chars?

Who is right here?

解决方案

§2.13.2/4:

The escape \xhhh consists of the backslash followed by x followed by one or more hexadecimal digits that are taken to specify the value of the desired character. There is no limit to the number of digits in a hexadecimal sequence. A sequence of octal or hexadecimal digits is terminated by the first character that is not an octal digit or a hexadecimal digit, respectively.

She is right.

However, you can terminate it early by eager catenation: the sequence of literals "\x000a" "axz" specifies a single four-character string literal. (2.13.4/3)

Also note that Unicode uses 21-bit code points; it doesn't stop at 16 bits.

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