>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?
>>
