Paul Herring wrote:
> On Tue, Mar 4, 2008 at 12:01 PM, debasish deka
> <[EMAIL PROTECTED]> wrote:
>> Hello All,
>> A few days back I attended Huawei Interview where the following code's
>> output was expected:
>> unsigned char i;
>> for(i=0;i<2000;i++)
>> printf("\t%d",i);
>> A simple thinking gives the output to be :
>> values from 0 to 255.
>> When I checked it out on my Dev-C++ it showed counting exactly up to 2000.
>
> You compiler isn't doing what it should then.
>
>> Now my doubt it if a character variable occupies one byte then it should
>> count upto 255 only
>
> On most computers this is what it should be doing. (The exceptions
> being when a char has more than 8 bits.)
>
>> then it should overflow. Thus the result should be either counting upto
>> 255(and stop)
>
> Erm - nope.
>
>> or ending in an infinite loop counting each time up to 255.
>
> That's what it should be doing.
>
>> Then why and how it is counting uoto 2000 (exactly).
>
> Pass. Your compiler may be broken.
>
>> Again in a forum it was asked to give the output of the following codes ::
>> unsigned i;
>> i=100*400;
>> printf("\n%d\n",i);
>> Strangely again, the output was exactly 40000as opposed to + 32767 or
>> showing
>> any sort or overflow, since the range of unsigned int ranges from- 32767 to
>> 32768.
>
> That is one possible range. The Standard dictates minimum ranges only
> - compilers are allowed to exceed them.
"unsigned int has a range from -32767 to 32768"?
OP: You might want to re-read that statement again. Particularly the
'unsigned' part.
>> Now my question is how the compiler manages to tackle such situations, I
>> mean what
>> mechanism does the compiler the aplies in such a situation...
>
> Depends on your compiler, and the value of CHAR_BIT (8 on my system.)
> I get the following:
Only thing I can think of is that Unicode-sized characters might be in
effect for simple char types? Finding out the value of CHAR_BIT is a
good idea.
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