Using octal character gives warning: multi-character character constant

Question

Following the logic from this question, the following code should work:

#include <stdio.h>

int main(){
    printf("%c", '\0101');
    return 0;
}

However, it gives the following error:

main.c: In function 'main':
main.c:5:18: warning: multi-character character constant [-Wmultichar]
     printf("%c", '\0101');
                  ^~~~~~~
exit status -1

I am not sure why it is a multi-character constant. I believe there should only be a single character constant inside those single quotes (octal 101 = decimal 65 = 'A'). Why are there more than one characters? And why isn't octal notation not working?

I'm no C expert, but octal character literals are 3-digits, aren't they? You don't specify the initial 0, that's just for numeric literals I think. — Lasse V. Karlsen, Jun 22 '18 at 14:38
Why the error is talking about `"%s"` while the code has `"%c"`? — Eugene Sh., Jun 22 '18 at 14:45
@LasseVågsætherKarlsen For character literals leading zero is not needed. It is needed for integer octal literals. See https://en.cppreference.com/w/cpp/language/escape ... update ah.. it's for C++.. but same idea — Eugene Sh., Jun 22 '18 at 14:46
@EugeneSh. Sorry, I copied the wrong code, that `%s` is a typo. Although if you run the `repl` file I linked to, it gives the same error I was talking about. Also, it seems that the cppreference link is the answer I needed. Could you please post an answer? Thanks! — Gaurang Tandon, Jun 22 '18 at 14:51
when writing an OCTAL character, use a leading 0 but do NOT use a leading '\' — user3629249, Jun 23 '18 at 15:40

Bathsheba · Answer 1 · 2018-06-22T15:32:17.973

The octal char notation ought to be of the form \abc where a, b, and c, are octal digits (i.e. in the inclusive range of 0 to 7).

Your has four digits, so the compiler will interpret it as \010 (maximal munch) followed by 1.

That's a multicharacter constant, rather like '12'. Like \abc, that has an int type but the value is implementation defined, and the conversion to c in printf will have implementation-defined behaviour. Your helpful compiler is alerting you of that and, not surprisingly, is using correct terminology.

Did you mean to write '\101'? If you did, and what you really wanted was the upper case letter A, then write 'A' for portable C.

score 6 · Accepted Answer · edited Jun 22 '18 at 14:55

6

It should be '\101' not '\0101'. You can use '\x41'(Hexadecimal) or '\101'(octal) instead of 'A'. But both reduces the portability and readability of your code. You should only consider using escape sequences when there isn't a better way to represent the character.

edited Jun 22 '18 at 14:55

Eugene Sh.

17,802
8
40
61

answered Jun 22 '18 at 14:52

anoopknr

3,177
2
23
33

1

Or you can use `'A'` for `'A'` :D – Eugene Sh. Jun 22 '18 at 14:52
@EugeneSh. that was just for demonstration purpose xD – Gaurang Tandon Jun 22 '18 at 14:53
Can you explain how it reduces _portability_ even if it is in standard? – Soner from The Ottoman Empire Jun 22 '18 at 15:14
2

@snr `'A'` gives you an `A` no matter what your machine's character set is. `'\101'` and the like assume ASCII. – Steve Summit Jun 22 '18 at 15:23
Unless you explicitly want to have `\101` regardless of how it is represented characterwise. – Eugene Sh. Jun 22 '18 at 15:25

Using octal character gives warning: multi-character character constant

2 Answers2

Linked