~ operator returns signed value

gcc is clearly returning -1 when I do this
~ (unsigned short) 0

which is not the case if "unsigned int" is used instead of "unsigned
short".

What does the C standard say about that? (reference to the standard
is appreciated.)

Thanks.

Reza.

re: ~ operator returns signed value

RezaRob wrote:If INT_MAX is greater than or equal to USHRT_MAX,
then (~ (unsigned short) 0) means the same thing as
(~ (int)(unsigned short) 0).



ISO/IEC 9899: 1990
6.2.1 Arithmetic operands
6.2.1.1 Characters and integers
A char, a short int, or an int bit-field,
or their signed or unsigned varieties,
or an enumeration type, may be used in an expression
wherever an int or unsigned int may be used.
If an int can represent all values of the original type,
the value is converted to an int; otherwise,
it is converted to an unsigned int.
These are called the integral romotions.
All other arithmetic types are unchanged by the integral promotions.

--
pete

re: ~ operator returns signed value

pete wrote:That should be "These are called the integral promotions."
--
pete

re: ~ operator returns signed value

On Sep 23, 3:47 pm, pete <pfil...@mindspring.comwrote:Pete, I really appreciate this.

Reza.

re: ~ operator returns signed value

RezaRob wrote:The following three expressions have the same type and value:
(USHRT_MAX)
((unsigned short) ~0u)
((unsigned short) -1)

--
pete

re: ~ operator returns signed value

Are you sure about the middle one?

1: 0u is of type unsigned int.

2: ~0u is of type unsigned int and is equal to UINT_MAX.

3: (unsigned short)~0u will be UINT_MAX % USHRT_MAX, or at least I
think it will, so it can't be USHRT_MAX.

but of course I'm open to correction!

Martin

re: ~ operator returns signed value

On Sep 24, 4:36 pm, Martin Wells <war...@eircom.netwrote:Sure, there's a cast in it to unsigned short.
;-)
[snip]

re: ~ operator returns signed value

Martin Wells said:
Oh good. :-) Re your Point 3, a single counter-example should suffice.
Consider a system with UINT_MAX == USHRT_MAX (e.g. typical 1990s MS-DOS
systems, or an SIL64 system such as at least one Cray. Clearly, on such a
system, UINT_MAX % USHRT_MAX will be 0 (for the same reason that x % x is
0), and yet (unsigned short)~0u will be equivalent to ~0u, which is most
emphatically NOT 0.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

re: ~ operator returns signed value

Richard:

Sorry I've red that a few times but I still don't understand what
you're saying.

Was pete right or wrong?

Martin

re: ~ operator returns signed value

Martin Wells said:
You claimed that "(unsigned short)~0u will be UINT_MAX % USHRT_MAX". I
disputed that claim, by pointing out a counter-example.
Presumably. :-)

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

re: ~ operator returns signed value

Richard:

I'm still lost. I'll try think it through using MS-DOS as an example:

1: 0u == the number 0
2: ~0u == UINT_MAX == the number 65535
3: (unsigned short)~0u == the number USHRT_MAX % 65535 == the
number 65535 % 65535 == the number 0.

This leads me to believe that pete was wrong... unless I'm missing
something.

Martin

re: ~ operator returns signed value

Martin Wells said:
<snip>
Please explain why you think this is the case. I can see no justification
for your assumption that (unsigned short)~0u is equal to USHRT_MAX % 65535
(and yes, I know we're talking 16-bit systems. You originally wrote that
it's equal to UINT_MAX % USHRT_MAX, which is also incorrect.

You might reasonably claim that, on a 16-bit system, (unsigned short)~0u is
equal to UINT_MAX % (USHRT_MAX + 1), but that would just be equal to
UINT_MAX (and indeed USHRT_MAX).
You appear to be missing a + 1. :-)

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

re: ~ operator returns signed value

Martin Wells wrote:This part of your post: "UINT_MAX % USHRT_MAX"
is wrong.

--
pete

re: ~ operator returns signed value

pete:

I'm gonna give the middle one another go. . .

Before I begin though, I'm gonna pretend I'm working with the
following machine:

CHAR_BIT == 11
sizeof(short) == 2 (all 22 bits are value bits)
sizeof(int) == 3 (31 bits are value bits, 1 is padding)

Therefore, we have:

UINT_MAX = 2147483647
USHRT_MAX = 4194303

1: 0u

Expression Type: unsigned int
Expression Value: 0

2: ~0u

Expression Type: unsigned int
Expression Value: 2147483647 (i.e. UINT_MAX)

3: (unsigned short)~0u

== UINT_MAX % (USHRT_MAX + 1)
== 2147483647 % 4194304
== 4194303

Oh, so that worked out OK. I suppose what I was tryna get my head
around from the start was how you could so blindy make the assumption

Given: a is a positive integer
b is a positive integer
a is greater than or equal to b
That:

( pow(2,a) - 1 ) % pow(2,b)

is equal to:

pow(2,b) - 1

Martin

re: ~ operator returns signed value

Shuda written "2 are padding" instead of "1 is".

Martin

re: ~ operator returns signed value

Martin Wells wrote:Actually, I got one of those wrong.
The type of USHRT_MAX isn't unsigned short.

--
pete