number of digits in a hex escape code in C/C++

Question

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?

Answer 1

§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.

Answer 2

Quote from MSDN on C++ character constants:

Octal escape sequences, specified in the form \ooo, consist of a backslash and one, two, or three octal characters. Hexadecimal escape sequences, specified in the form \xhhh, consist of the characters \x followed by a sequence of hexadecimal digits. Unlike octal escape constants, there is no limit on the number of hexadecimal digits in an escape sequence.

from http://msdn.microsoft.com/en-us/library/6aw8xdf2.aspx