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why does printf get -1 for an unsigned integer?


#include <stdio.h>

int main(void)
{
unsigned int u = 0;

u--;
if (u < 0)
printf("u < 0\n");
else
printf("u >= 0\n");
printf("u: %d\n", u);
return 0;
}

$ cc a.c
$ ./a.out
u >= 0
u: -1
$

On 11 Apr, 10:42, "lovecreatesbea...@gmail.com"
<lovecreatesbea...@gmail.com> wrote:
> why does printf get -1 for an unsigned integer?

It doesn't know how to interpret what you've given it, expect by the
"mask" you specify.

> #include <stdio.h>
>
> int main(void)
> {
> unsigned int u = 0;
>
> u--;
[snip]
> printf("u: %d\n", u);
[snip]

You lied to printf.

Try reading the manual before posting this sort of nonsense.

said:

> why does printf get -1 for an unsigned integer?

Because you tricked printf. You said you were giving it a signed int,
and it believed you.

Use %u, not %d, to print unsigned ints.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.

"" <> writes:
> why does printf get -1 for an unsigned integer?
>
>
> #include <stdio.h>
>
> int main(void)
> {
> unsigned int u = 0;
>
> u--;
> if (u < 0)
> printf("u < 0\n");
> else
> printf("u >= 0\n");
> printf("u: %d\n", u);
> return 0;
> }
>
> $ cc a.c
> $ ./a.out
> u >= 0
> u: -1
> $

Because you lied to it. "%d" expects an argument of type int; you
gave it an argument of type unsigned int. Try "%u".

(You're probably seeing that behavior because (unsigned int)-1, or
UINT_MAX, has the same representation as (int)-1 in 2's-complement,
but that's not guaranteed.)

--
Keith Thompson (The_Other_Keith) kst- <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

On Apr 11, 3:04 am, mark_blue...@pobox.com wrote:
> On 11 Apr, 10:42, "lovecreatesbea...@gmail.com"
>
> <lovecreatesbea...@gmail.com> wrote:
> > why does printf get -1 for an unsigned integer?
>
> It doesn't know how to interpret what you've given it, expect by the
> "mask" you specify.
>
> > #include <stdio.h>
>
> > int main(void)
> > {
> > unsigned int u = 0;
>
> > u--;
> [snip]
> > printf("u: %d\n", u);
>
> [snip]
>
> You lied to printf.
>
> Try reading the manual before posting this sort of nonsense.

Yes Minister

wrote:
> why does printf get -1 for an unsigned integer?
>
>
> #include <stdio.h>
>
> int main(void)
> {
> unsigned int u = 0;
>
> u--;
> if (u < 0)

Because u is unsigned, the above condition is never satisfied

> printf("u < 0\n");
> else
> printf("u >= 0\n");
> printf("u: %d\n", u);

Because u is unsigned, printing its value with "%d", specifying a
_signed_ value is an error.

> return 0;
> }
>
> $ cc a.c
> $ ./a.out
> u >= 0
> u: -1
> $
>