big decimal integer constant warning

[fctk]

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

i read both here
(http://www-ccs.ucsd.edu/c/express.html#decimal integer constant) 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 integer constant) 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.

 

[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.

 

[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.

That's because the rules changed in 1999.

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]

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...

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