big decimal integer constant warning
fctk
i'm trying to compile this small program:
int main(void) {
unsigned long int max;
max = 4000000000;
return 0;
}
but i get the following warning, and can't understand why:
limits.c: In function `main':
limits.c:5: warning: this decimal constant is unsigned only in ISO C90
i read both here
(http://www-ccs.ucsd.edu/c/express.html#decimal%20integer%20constant)
and on my book (Kelley-Pohl), that if i have a decimal integer constant
with no suffix, then the compiler will choose among the following
ordered list:
1) int
2) long int
3) unsigned long int
the first one which is able to represent that constant. since on my
system both int and long int are 4 bytes sized, then i would expect
4000000000 is unsigned long int. since i am assigning an unsigned long
int constant to an unsigned long int variable, i don't see why i should
get that warning.
i know two ways to avoid the warning:
1) specifing 4000000000UL instead of 4000000000
2) using 0xEE6B2800 instead of 4000000000
but i'd like to understand why 4000000000 is not ok.
note that i'd like to write only ANSI/ISO C compatible programs
(C89/C90), no traditional C or C99 ones (after having learned C89, i
will switch to C99 if i'll have time).
i am using gcc version 3.4.4 on a gentoo linux system.
parameters: gcc -x c -ansi -pedantic -Wall -Wextra (i get that warning
also with no parameters at all)
also, i found this parameter:
-Wtraditional
[...]
* The ISO type of an integer constant has a different width or
signedness from its traditional type. This warning is only issued if
the base of the constant is ten. I.e. hexadecimal or octal values,
which typically represent bit patterns, are not warned about.
[...]
anyway, even if i put -Wno-traditional, i always get that warning.
Eric Sosman
fctk wrote On 03/29/06 10:26,:
> hello,
>
> i'm trying to compile this small program:
>
> int main(void) {
>
> unsigned long int max;
>
> max = 4000000000;
>
> return 0;
>
> }
>
> but i get the following warning, and can't understand why:
>
> limits.c: In function `main':
> limits.c:5: warning: this decimal constant is unsigned only in ISO C90
The compiler is warning you that the rules changed
between C90 and C99, and that the type of the constant
depends on which Standard you follow -- to put it another
way, the type of the constant is "unstable" in the sense
that it could change with a compiler upgrade.
For C90, the rule is as you stated: the type will be
the first of int, long int, or unsigned long int that
can accommodate the value, and on your system this comes
down to unsigned long int.
For C99, the type will be the first of int, long int,
or long long int that can accommodate the value, which
for your system would turn out to be long long int.
Recommendation: If you want an unsigned long, tack
on a "UL" suffix.
Jack Klein
> hello,
>
> i'm trying to compile this small program:
>
> int main(void) {
>
> unsigned long int max;
>
> max = 4000000000;
>
> return 0;
>
> }
>
> but i get the following warning, and can't understand why:
>
> limits.c: In function `main':
> limits.c:5: warning: this decimal constant is unsigned only in ISO C90
That's because the rules changed in 1999.
> i read both here
> (http://www-ccs.ucsd.edu/c/express.html#decimal%20integer%20constant)
> and on my book (Kelley-Pohl), that if i have a decimal integer constant
> with no suffix, then the compiler will choose among the following
> ordered list:
>
> 1) int
> 2) long int
> 3) unsigned long int
>
> the first one which is able to represent that constant. since on my
> system both int and long int are 4 bytes sized, then i would expect
> 4000000000 is unsigned long int. since i am assigning an unsigned long
> int constant to an unsigned long int variable, i don't see why i should
> get that warning.
That's because the rules changed in 1999. If your compiler uses the
C99 rules, an unadorned decimal constant is the first one of:
-- signed int
-- signed long
-- signed long long
...in which it fits. Under C99, an unadorned decimal constant never
becomes an unsigned type.
So the diagnostic is telling you that the constant is interpreted
differently depending on which version of ISO C is being conformed to.
If the compiler is invoked in C90 conforming mode, the type of the
constant is "unsigned long", but in C99 conforming mode, the type is
"signed long long". If the destination was a signed type, the actual
value after initialization might be different between the two
versions. And this code could break on other platforms.
> i know two ways to avoid the warning:
>
> 1) specifing 4000000000UL instead of 4000000000
> 2) using 0xEE6B2800 instead of 4000000000
>
> but i'd like to understand why 4000000000 is not ok.
Nobody says that it isn't "OK". A compiler is allowed to issue a
diagnostic about anything for any reason, as long as it issues those
required by the standard.
> note that i'd like to write only ANSI/ISO C compatible programs
> (C89/C90), no traditional C or C99 ones (after having learned C89, i
> will switch to C99 if i'll have time).
Then follow the MISRA rule (one of the good ones) about always
specifying a type for numeric literals.
> i am using gcc version 3.4.4 on a gentoo linux system.
> parameters: gcc -x c -ansi -pedantic -Wall -Wextra (i get that warning
> also with no parameters at all)
IIRC, 3.x versions of gcc had some C99 features included, and they,
and even earlier versions, of gcc supported the "long long" int types
as an extension to C90.
> also, i found this parameter:
>
> -Wtraditional
>
> [...]
>
> * The ISO type of an integer constant has a different width or
> signedness from its traditional type. This warning is only issued if
> the base of the constant is ten. I.e. hexadecimal or octal values,
> which typically represent bit patterns, are not warned about.
>
> [...]
>
> anyway, even if i put -Wno-traditional, i always get that warning.
The real answer is "don't do this". Specify the exact type of your
literal with a suffix, for maximum portability under any version of
the C standard.
Also consult your compiler documentation. There is some way to tell
gcc which version of the C standard you want it to conform to.
--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://c-faq.com/
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~ajo/docs/FAQ-acllc.html
Clark S. Cox III
> hello,
>
> i'm trying to compile this small program:
>
> int main(void) {
>
> unsigned long int max;
>
> max = 4000000000;
>
> return 0;
>
> }
>
> but i get the following warning, and can't understand why:
>
> limits.c: In function `main':
> limits.c:5: warning: this decimal constant is unsigned only in ISO C90
Because under C99, the type of the constant is "long long" (which is signed),
while under C90 (assuming 4 byte ints and longs), it is "unsigned long" (which
is unsigned). Hence, the warning is telling you that your constant is unsigned
under C90, but not under C99, and that you may get different results when you
change compilers.
--
Clark S. Cox, III
clar...@gmail.com
Spoon
> Also consult your compiler documentation. There is some way to tell
> gcc which version of the C standard you want it to conform to.
http://gcc.gnu.org/onlinedocs/gcc-3.4.6/gcc/C-Dialect-Options.html
-std=c89 (same as -ansi)
-std=c99
fctk
> The compiler is warning you that the rules changed
> between C90 and C99, and that the type of the constant
> depends on which Standard you follow -- to put it another
> way, the type of the constant is "unstable" in the sense
> that it could change with a compiler upgrade.
Jack Klein wrote:
> That's because the rules changed in 1999.
Clark S. Cox III wrote:
> Because under C99, the type of the constant is "long long" (which is signed),
> while under C90 (assuming 4 byte ints and longs), it is "unsigned long" (which
> is unsigned). Hence, the warning is telling you that your constant is unsigned
> under C90, but not under C99, and that you may get different results when you
> change compilers.
ok, it seems you all agree on the reason of that warning.
i tried to specify -std=iso9899:1990 (C89/C90) in the gcc parameters
list, but i get that warning again.
i still wonder. i don't mind about C99, i'd like gcc not to be aware of
its existence. when C89/C90 was made (1989/1990) C99 didn't existed, so
why telling me my program can produce different results if compiled with
C99 standard? C99 doesn't exist...
i hope it's clear what i mean...
Michael Mair
Yes.
This is rather a request to the compiler to not warn you about
problems other people might have on other compilers when
compiling your code.
Note that, in principle, a compiler can warn you about everything
it likes to ("Your code contains 42 'i's -- that is a bad omen on
days like this one").
Maybe, in gnu.gcc.help someone can tell you how to achieve that
(or maybe you find something in the gcc documentation yourself).
However, if you go this way, then consider at least protecting
these others (or maybe even yourself) from errors:
#if __STDC_VERSION__ >= 199901L
# error My code does not work for C99 or newer standards
#endif
If you are maintaining old code and if this warning comes up at
hundreds of different places, switching off or filtering this
warning and putting the above in one source file may be a
sensible way to go. The decision should be clearly documented
somewhere.
If you are writing new code, then the warning is justified and
removing its cause as often as it comes up is just sensible --
why leave unnecessary and potentially dangerous degrees of
freedom in your code if you can say exactly what you mean?
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.