what's the effect of uppercase "S" in printf

[marsarden]

code below:

#include <stdio.h>

int main(void)
{
char *aaa[]={"a","b","c"};
printf("%S\n",*aaa); /*the uppercase character S*/
getchar();

}

the stdout give me : "abc"
but ,if i change the "%S" to "%s" /*lowercase s*/
it give me "a" only.
there are no info about %S in c std lib reference,
any one can tell me the truth ?

[Spiros Bousbouras]

marsarden wrote:

code below:

#include <stdio.h>

int main(void)
{
char *aaa[]={"a","b","c"};
printf("%S\n",*aaa); /*the uppercase character S*/
getchar();

}

the stdout give me : "abc"
but ,if i change the "%S" to "%s" /*lowercase s*/
it give me "a" only.
there are no info about %S in c std lib reference,
any one can tell me the truth ?

The S conversion specifier comes from SUSv2 and is
not part of standard C. It has the same meaning as the
ls specifier which is that the argument is of type wchar_t.

I only have a hazy understanding of wide characters so
I can't give you a full explanation why you get output abc
when using the S conversion. The answer may be implementation
specific. It must have something to do with the printf function
interpreting the bytes 'a' , NUL as part of a single wide character
so it doesn't terminate the wide character string there.

[Keith Thompson]

"marsarden" <ar*******@gmail.comwrites:

code below:

#include <stdio.h>

int main(void)
{
char *aaa[]={"a","b","c"};
printf("%S\n",*aaa); /*the uppercase character S*/
getchar();

}

the stdout give me : "abc"
but ,if i change the "%S" to "%s" /*lowercase s*/
it give me "a" only.
there are no info about %S in c std lib reference,
any one can tell me the truth ?

There is no "%S" format in standard C. As far as the standard is
concerned, it can do anything.

<OT>
On one system, the documentation for printf() says:

S (Not in C99, but in SUSv2.) Synonym for ls. Don't use.

"%ls" expects a wchar_t* argument, so again, you're getting undefined
behavior.
</OT>

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Rod Pemberton]

"Spiros Bousbouras" <sp****@gmail.comwrote in message
news:11*********************@i42g2000cwa.googlegro ups.com...

marsarden wrote:

code below:

#include <stdio.h>

int main(void)
{
char *aaa[]={"a","b","c"};
printf("%S\n",*aaa); /*the uppercase character S*/
getchar();

}

the stdout give me : "abc"
but ,if i change the "%S" to "%s" /*lowercase s*/
it give me "a" only.
there are no info about %S in c std lib reference,
any one can tell me the truth ?

The S conversion specifier comes from SUSv2 and is
not part of standard C. It has the same meaning as the
ls specifier which is that the argument is of type wchar_t.

I only have a hazy understanding of wide characters so
I can't give you a full explanation why you get output abc
when using the S conversion. The answer may be implementation
specific. It must have something to do with the printf function
interpreting the bytes 'a' , NUL as part of a single wide character
so it doesn't terminate the wide character string there.

References:

GCC:
http://www.gnu.org/software/libc/man...%20Conversions
" %S This is an alias for %ls which is supported for compatibility with the
Unix standard. "

Posix:
http://www.opengroup.org/onlinepubs/...ns/printf.html
1) "S [XSI] Equivalent to ls. "
2) "l (ell)
Specifies that a following d, i, o, u, x, or X conversion specifier applies
to a long or unsigned long argument; that a following n conversion specifier
applies to a pointer to a long argument; that a following c conversion
specifier applies to a wint_t argument; that a following s conversion
specifier applies to a pointer to a wchar_t argument; or has no effect on a
following a, A, e, E, f, F, g, or G conversion specifier. "
Rod Pemberton

[Richard Heathfield]

marsarden said:

code below:

#include <stdio.h>

int main(void)
{
char *aaa[]={"a","b","c"};
printf("%S\n",*aaa); /*the uppercase character S*/
getchar();

}

the stdout give me : "abc"

The C language does not make any provision for what happens when you tell
printf to expect a format specifier of a kind it understands, and then fail
to provide one. So the behaviour of your program is undefined, and it could
do anything at all, including (but by no means restricted to) what you say
it did do.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)