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Re: [vox-tech] gcc warning question
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Re: [vox-tech] gcc warning question



On Tue, 15 Oct 2002, Peter Jay Salzman wrote:

> hi sam,
> 
> i wasn't really worrying about fitting numbers into correct datatypes as
> much as i was the warning that my "unsigned int" was getting truncated
> to an unsigned int.
> 
> seems like an odd thing for gcc to say.

It isn't complaining about moving the data from the constant to the
variable.  It is complaining about your attempt to describe 99999999999 as
an unsigned integer by sticking a "u" after it.  Within the definition
supported by gcc, it doesn't qualify.  Thus, the four bytes stuffed into
the code image to be copied into the stack upon execution would not
represent the number your source code contained, even though no conversion
is taking place at runtime.

> 
> pete
> 
> begin Samuel Merritt <spam@andcheese.org> 
> > On Tue, Oct 15, 2002 at 11:55:12AM -0700, Peter Jay Salzman wrote:
> > > gcc warns:
> > > 
> > >    % .!gcc -W -Wall test.c
> > >    test.c: In function `main':
> > >    test.c:6: warning: large integer implicitly truncated to unsigned type
> > > 
> > > the variable was declared as an unsigned and the rvalue has the unsigned
> > > suffix.  what's wrong with this code?   why is gcc complaining? 
> > 
> > I think you need to use an unsigned long. It looks like sizeof(unsigned
> > int) = 4, and 99999999999 > 2^32. 
> >  
> > > pete
> > > 
> > > 
> > > 
> > > #include <stdio.h>
> > > 
> > > int main(void)
> > > {
> > > 	unsigned int size = 99999999999u;
> > > 
> > > 	printf("size is %u.\n", size);
> > > 
> > > 	return 0;
> > > }

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