RE: unsigned

["Young, Michael" <Michael dot Young at paetec dot com>]

>C tries to generate integers by default, so when you write that number,
>which is larger than the signed integer boundary, it will instead
>generate an unsigned integer instead and let you know about it. You
>should be able to get rid of the warning by writing it as 2147483648U (U
>for unsigned).
>

Yes.  See C++ Std 2.13.1 for more info on this.  However, I'd like to point
out that the standard doesn't guarantee that 2147483648 is representable by
an integer or that an integer is 4 bytes in length - that is implementation
dependent (really, CPU/platform dependent - see 3.9.1).  On a 16-bit machine,
with a compiler that has 4 byte long integers, you may need to write "2147483648UL".
You didn't mention the platform, or the exact compiler you're using - I assumed
g++, but, from the extensions in the example below, maybe it was just the C compiler.

But I have a question - how does one write this portably (in C++)?

   typedef unsigned int uint_32_t ;  // similar to C99 types and common practice
   ...
   uint_32_t squiggy( 2147483648 ) ;  // Does this work w/o generating a warning?
   ...

if the above code still generates a warning, then does one resort to the following? :

   // Macro definition changes with platform, depending on length of integer
   #define UINT32_LITERAL( x ) xUL
   ...
   uint_32_t squiggy( UINT32_LITERAL( 2147483648 ) ) ;  
   ...
 
>On 12/04/2006 03:48 PM, Trevis Rothwell wrote:
>> Given the following program:
>> 
>>  int main()
>>  {
>>    unsigned int squiggy = 2147483648;
>>  }
>> 
>> Compliing on GCC 3.2.3, I get the following warning:
>> 
>>  $ gcc foo.c
>>  foo.c: In function `main':
>>  foo.c:3: warning: decimal constant is so large that it is unsigned
>> 
>> To the best of my knowledge, 2147483648 should fit with ample room to
>> spare in an unsigned (4-byte) integer.
>> 
>> What's going on?
>>