what is the reason ?

[codergem]

It will not run. This is fine.
main()
{
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}

but in the same code if i ll make unsigned int as short unsigned int
then it will run for an infinite loop.
what is the reason?
thanks.

 

[Igmar Palsenberg]

It will not run. This is fine.
main()
{
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}

but in the same code if i ll make unsigned int as short unsigned int
then it will run for an infinite loop.
what is the reason?

Turn on compiler warnings and you can probably figure this out yourself.


Igmar

 

[Eric Sosman]

It will not run. This is fine.
main()
{
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}

but in the same code if i ll make unsigned int as short unsigned int
then it will run for an infinite loop.
what is the reason?

The "usual arithmetic conversions."

 

[codergem]

Sorry Man
I dont know how turn ON this?

 

[Clever Monkey]

Sorry Man
I dont know how turn ON this?

"Consult your documentation."

For example, if I was using Microsoft VC, I would consult
http://msdn.microsoft.com or the local help files. For GCC, similar
help is available.

See also, <http://catb.org/~esr/faqs/smart-questions.html>

 

[Ancient_Hacker]

well in general you're hoping that an unsigned int will somehow count
down to -2.

Unlikely on the face of it, but you say it works fine.


That may be, but only due to a binary coincidence.


If you change it to a short, then it may seem to wrap around funny,
like from 2, 1, 0, -65535.

In any case, if you listen to the compiler warnings, you wouldnt have
this problem.

Styrongly suggest yuou use -warn=99 or whatever. "C" is hard enough--
you can't afford to ignore the few warnigns the compiler can put out.

 

[Flash Gordon]

Ancient_Hacker wrote:

Please provide context. You've been here long enough to know it is expected.

The post you were replying to said

| It will not run. This is fine.
| main()
| {
| unsigned int i;
| for(i=1;i>-2;i--)
| printf("c aptitude");
| }
|
| but in the same code if i ll make unsigned int as short unsigned int
| then it will run for an infinite loop.
| what is the reason?
| thanks.

well in general you're hoping that an unsigned int will somehow count
down to -2.

Unlikely on the face of it, but you say it works fine.


That may be, but only due to a binary coincidence.

Or it could be because it does exactly what the C standard requires it
does. Look up the usual arithmetic promotions.

If you change it to a short, then it may seem to wrap around funny,
like from 2, 1, 0, -65535.

I somehow doubt it will wrap from 0 to -65535 under any conditions ;-)

In any case, if you listen to the compiler warnings, you wouldnt have
this problem.

Styrongly suggest yuou use -warn=99 or whatever. "C" is hard enough--
you can't afford to ignore the few warnigns the compiler can put out.

This advice is good. For details of how to get a compiler to produce
decent warnings you are best asking on a group dedicated to your
compiler. Or reading the manual/help, of course!

 

[Peter Nilsson]

Turn on compiler warnings and you can probably figure this out yourself.

Despite being riddled with issues, there is nothing in the code that
requires a
C90 compiler to issue a diagnostic. I know of at least 2 compilers that
woundn't
issue the particular desired warning, no matter how high you turned
them up.

IIRC, the gcc diagnostic warns about the cause, but doesn't elaborate
on the
issue. So, given the demonstration of coding so far, I seriously doubt
that the
OP will figure it out.

Fortunately, the problem (actually its mirror) is covered in the FAQ...

http://www.c-faq.com/expr/preservingrules.html