length of string containing \0
char p[]="abc\012\0x34";
printf("%d\n",strlen(p));
I 开发者_开发技巧am getting output 4. Shouldn't it be 3 ??? Although for following i am getting 3.
char p[]="abc\0";
printf("%d\n",strlen(p));
Your string does contain four characters before the \0
, i.e. abc
and \012
.
The latter is a valid octal escape sequence, which is 10 in decimal, i.e an ASCII linefeed character.
\0x34
on the other hand isn't valid octal - only the \0
part is valid hence that's the real end of your NUL terminated string.
\012
is an octal escaped character, not a NUL
followed by 1
and 2
. x
terminates the second octal character so it is genuinely a NUL
. (\x34
would be the correct form for a hexadecimal escaped character.)
The representation of a NUL
character as \0
is just a special case of an octal escape sequence. In general a \
can be followed by one, two or three octal digits to form a valid octal escape sequence in a character or string literal.
Your string has length 4:
You code is equivalent to: char p[]={'a','b'.'c'.'\012','\0','x','3','4','\0'};
\012 - character with code 12 in octal numeral system (= 10 in decimal = '\n')
\012
is a single character. It's stopping on the \0
after that (and "x34"
is three more characters, not counting the NUL terminator).
\012
is an octal value("\n").
精彩评论