Using C am I right in thinking that literals beginning with multiple zeros are considered octal?

Question

In the following C code are octal literals used for all these defines? Even if they start with multiple zeros?

#define TCL_REG_BASIC       000000  /* BREs (convenience). */
#define TCL_REG_EXTENDED    000001  /* EREs. */
#define TCL_REG_ADVF        000002  /* Advanced features in EREs. */
#define TCL_REG_ADVANCED    000003  /* AREs (which are also EREs). */
#define TCL_REG_QUOTE       000004  /* No special characters, none. */
#define TCL_REG_NOCASE      000010  /* Ignore case. */
#define TCL_REG_NOSUB       000020  /* Don't care about subexpressions. */
#define TCL_REG_EXPANDED    000040  /* Expanded format, white space & comments. */
#define TCL_REG_NLSTOP      000100  /* \n doesn't match . or [^ ] */
#define TCL_REG_NLANCH      000200  /* ^ matches after \n, $ before. */
#define TCL_REG_NEWLINE     000300  /* Newlines are line terminators. */
#define TCL_REG_CANMATCH    001000  /* Report details on partial/limited * matches. */

Yes, that's correct, though it only takes the one leading zero. — doynax, Dec 16 '13 at 21:37
**especially** if they start with multiple zeros ;) (hex literals start with 0x) — Tim Seguine, Dec 16 '13 at 21:38
The author is probably trying to hint at the width of the bitfield. And wants them to all line up nicely — Tim Seguine, Dec 16 '13 at 21:40
What Tim said. This kind of thing is pretty standard (used all over the place in windows headers). — Taylor Brandstetter, Dec 16 '13 at 21:42
Idea for an esoteric programming language: Make numbers with one leading zero octal, two leading zeros decimal again, three leading zeros dozenal, four leading zeros base-14, five leading zeros hexadecimal, etc. — dan04, Dec 16 '13 at 21:46

score 4 · Accepted Answer · answered Dec 16 '13 at 21:40

4

From C Standard, 6.4.4.1 Paragraph 3:

An octal constant consists of the preﬁx 0 optionally followed by a sequence of the digits 0 through 7 only

answered Dec 16 '13 at 21:40

Taylor Brandstetter

3,523
15
24

When I posted my answer your's was not here! – haccks Dec 16 '13 at 22:00
1

@haccks I removed it when I saw that paulsm4 had already answered, but then someone asked me to add it back... Sorry for the confusion. :P – Taylor Brandstetter Dec 16 '13 at 22:14

score 4 · Answer 2 · edited Jun 20 '20 at 09:12

4

Yes You are right.

C11, 6.4.4.1 Integer constants:

An octal constant consists of the prefix 0 optionally followed by a sequence of the digits 0 through 7 only.

edited Jun 20 '20 at 09:12

Community

1
1

answered Dec 16 '13 at 21:42

haccks

104,019
25
176
264

Out of curiosity what disallows decimals literals with leading zeroes? E.g. 08 is a syntax error as far as GCC and MSVC is concerned. – doynax Dec 16 '13 at 21:48
@doynax; Since when `8` comes to the octal representation of a number? And why compiler should not give the warning? I suggest you to read number system. – haccks Dec 16 '13 at 21:54
I was just curious what the rule identifying 08 as an invalid octal number rather than a proper decimal literal might be since you seemed to have the standard handy. No real reason, just the inner lawyer in me wanting to pick holes in any formal definitions. – doynax Dec 16 '13 at 22:01
@doynax the same section 6.4.4.1 of the standard specifically defines a decimal constant as starting with a non-zero decimal digit. – Nigel Harper Dec 16 '13 at 22:03
2

@Nigel Harper: Damn. And I suppose the sequence of 0-7 digits here may be empty, thus covering the "0" case as a (very common) octal literal – doynax Dec 16 '13 at 22:06
I am pissed off by downvoting without any explanation/comment. This is cowardice. – haccks Dec 16 '13 at 22:07
@doynax Exactly - the "optionally followed" bit in the standard quote means that a bare 0 is technically a valid octal constant. – Nigel Harper Dec 16 '13 at 22:12
1

@doynax; Here is the complete para: `A decimal constant begins with a nonzero digit and consists of a sequence of decimal digits. An octal constant consists of the prefix 0 optionally followed by a sequence of the digits 0 through 7 only. A hexadecimal constant consists of the prefix 0x or 0X followed by a sequence of the decimal digits and the letters a (or A) through f (or F) with values 10 through 15 respectively.` Which clearly suggest that a decimal number can't begins with `0`. Continued........ – haccks Dec 16 '13 at 22:13
@doynax; .....This means that [a bare `0` is not a decimal integer but an octal integer](http://stackoverflow.com/a/6895543/2455888). – haccks Dec 17 '13 at 12:44
@haccks: Indeed, and Nigel concurs. Thankfully octal numbers have no other properties compared to decimal ones (aside from unsigned promotion) and are thus indistinguishable in this case as near as I can tell – doynax Dec 17 '13 at 13:09
@doynax; I commented regarding your comment:`I was just curious what the rule identifying 08 as an invalid octal number rather than a proper decimal literal might be since you seemed to have the standard handy. No real reason, just the inner lawyer in me wanting to pick holes in any formal definitions` – haccks Dec 17 '13 at 13:11

score 0 · Answer 3 · answered Dec 16 '13 at 21:38

Yes.

http://www.cplusplus.com/doc/tutorial/constants/

In addition to decimal numbers (those that most of us use every day), C++ allows the use of octal numbers (base 8) and hexadecimal numbers (base 16) as literal constants. For octal literals, the digits are preceded with a 0 (zero) character. And for hexadecimal, they are preceded by the characters 0x (zero, x). For example, the following literal constants are all equivalent to each other:

75         // decimal
0113       // octal
0x4b       // hexadecimal

Technically the question is about C, but thankfully C and C++ are the same in this regard — Taylor Brandstetter, Dec 16 '13 at 21:41

score 0 · Answer 4 · answered Dec 16 '13 at 23:57

OP's examples are all octal as they begin with a 0. Multiple leading zeros or the presence of an 8 or 9 does not change it away from being octal. An 8 or 9 should case an error.

FYI: Some additional occurrences (though maybe not a literal) in C where decimal, octal and hexadecimal interpretation occur.

Although not a literal, in a format string for printf(), specifiers like "%016llx" are not an octal 016 width, but a flag '0' and a decimal value of 16.

With printf(), output using specifier "%a", the output is in the style "[−]0xh.hhhh p±d," where the significand is in hexadecimal and the exponent is in decimal.

In escape sequences there is no decimal specification. Some samples follow:

\' \" \? \\ \a \b \f \n \r \t \v
\0   (octal)
\01   (octal)
\012   (octal)
\0123   (bad - only up to 3)
\x0 (hexadecimal)
\x01 (hexadecimal)
\x012 (hexadecimal)
\x0123 (hexadecimal)
\x01234… (hexadecimal)
\u1234 (hexadecimal)
\U00012345 (hexadecimal)

Using C am I right in thinking that literals beginning with multiple zeros are considered octal?

4 Answers4

C11, 6.4.4.1 Integer constants:

Linked

Related