Why does a C null terminator `\0` show up as `\000` during GDB debugging?

Question

During my GDB debugging sessions, I've noticed that null terminator characters, denoting the end of a string, and shown as \0 in C files, show up as \000 in GDB when displaying the value of a variable storing such a character.

(gdb) print buffer[10]
$2 = 0 '\000'

Can anyone tell me why that is?

Answer 1

GDB seems to always use 3 octal digits to display character escapes - and for a good reason_ Consider the following string

const char *str = "\1\2\3\4\5";

then

(gdb) p str
$1 = 0x555555556004 "\001\002\003\004\005"

This is because C standard says that an escape sequence consists of maximum of 3 octal digits. Thus if you write:

"\0a"

it means string literal of two characters - null followed by a. But if you write

"\01"

it means a string literal of one character: ASCII code 1 - Start-of-Header control character. In fact the shortest way to write ASCII null followed by the digit 1 (i.e. ASCII code 49) in a string literal is "\0001" The other possibilities are "\0" "1" using string concatenation; separate escapes "\0\61"; or using hex escapes \x..., all of which will be even longer....

So by always using 3 octal digits, GDB can produce consistent output for strings - such that when copied to a C program will result in the same string during runtime. Furthermore the output routine is simpler because it does not need to consider the following character.

Answer 2

This record '\0' is an octal escape sequence of a character constant (literal).

An octal escape sequence may contain at most three octal digits.