compiler warning about literal

compiler warning about literal

[Luchezar Belev]

hi,

when i try to compile this c code:
long long t = 2684354560;
gcc (version 4.8.2) says: " ...warning: this decimal constant is
unsigned only in ISO C90 [enabled by default]"

i added the option -std=C90 but i still get the same warning.
Is there a way one can get rid of this warning (other than adding the
"u" suffix to the literal, which i can't do for complicated reasons
that are beyond this question)

I wonder why does gcc give such warning at all given that in either
standard the literal decimal number  "2684354560" is valid:
 a) in ISO C90 it will be interpreted as unsigned long and
 b) in ISO C99 it will be interpreted as signed long long

Re: compiler warning about literal

[Luchezar Belev]

note the following strange gcc behavior:

 "long long t = 268435456;" =>doesn't give warning
 "long long t = 2684354560;" => gives warning!
 "long long t = 26843545600;" => doesn't give warning!

it looks to me that there is something wrong here

Re: compiler warning about literal

[Nicholas Mc Guire]

On Mon, 17 Nov 2014, Luchezar Belev wrote:

> note the following strange gcc behavior:
> 
>  "long long t = 268435456;" =>doesn't give warning
>  "long long t = 2684354560;" => gives warning!
>  "long long t = 26843545600;" => doesn't give warning!
> 
> it looks to me that there is something wrong here

I think thats correct

268435456 fit and was positive 
2684354560 is a negative long on the right side and was sign extended -> warning
26843545600 is a long long I guess and thus again positive -> no warning

thx!
hofrat

Re: compiler warning about literal

[Luchezar Belev]

I think i got it why GCC gives the warning with -std=C90.
It basically wants to say:
"Now that we are under the C90 rules, this literal is 'unsigned long',
but under some newer standard it sould be 'signed long long', so make
sure your code won't broke if you try to compile it under new
standard".

Now the only question that remains is: why there is no way to disable
the warning?

Re: compiler warning about literal

[Luchezar Belev]

The reason i can't use suffixes is because the real code that bothers
me is a large automatically generated array of unsigned decimal
numbers (all are less than 2^32) and it isn't practical to edit them
by hand.
Also I can't easily change the tool that generates them to make it
place suffixes.

On Mon, Nov 17, 2014 at 7:28 PM, Luchezar Belev <lukcho@gmail.com> wrote:
> I think i got it why GCC gives the warning with -std=C90.
> It basically wants to say:
> "Now that we are under the C90 rules, this literal is 'unsigned long',
> but under some newer standard it sould be 'signed long long', so make
> sure your code won't broke if you try to compile it under new
> standard".
>
> Now the only question that remains is: why there is no way to disable
> the warning?

Re: compiler warning about literal

[Jonathan Wakely]

On 17 November 2014 16:22, Luchezar Belev <lukcho@gmail.com> wrote:
> hi,
>
> when i try to compile this c code:
> long long t = 2684354560;
> gcc (version 4.8.2) says: " ...warning: this decimal constant is
> unsigned only in ISO C90 [enabled by default]"
>
> i added the option -std=C90 but i still get the same warning.
> Is there a way one can get rid of this warning (other than adding the
> "u" suffix to the literal, which i can't do for complicated reasons
> that are beyond this question)

The right thing to do is use an ll suffix.

long long t = 2684354560ll;



> I wonder why does gcc give such warning at all given that in either
> standard the literal decimal number  "2684354560" is valid:
>  a) in ISO C90 it will be interpreted as unsigned long and
>  b) in ISO C99 it will be interpreted as signed long long

The fact it means different things in different cases is why you get a warning.