Velocity Reviews

Velocity Reviews (http://www.velocityreviews.com/forums/index.php)
-   C Programming (http://www.velocityreviews.com/forums/f42-c-programming.html)
-   -   Unsigned & signed int (http://www.velocityreviews.com/forums/t541494-unsigned-and-signed-int.html)

Kislay 10-02-2007 06:34 PM

Unsigned & signed int
 
Consider the following code snippet

unsigned int i=10;
int j= - 2; // minus 2
if(i>j)
cout<<"i is greater";
else
cout<<"j is greater";

Since i is unsigned , j is greater . I know why , but vaguely . Can
someone please explain the mechanics behind it . Is the unsigned int
converted to signed or is it vice-versa . Also , the same rules would
apply to char as well , right ?


Joachim Schmitz 10-02-2007 06:50 PM

Re: Unsigned & signed int
 
"Kislay" <kislaychandra@gmail.com> schrieb im Newsbeitrag
news:1191350059.674272.186400@22g2000hsm.googlegro ups.com...
> Consider the following code snippet
>
> unsigned int i=10;
> int j= - 2; // minus 2
> if(i>j)
> cout<<"i is greater";
> else
> cout<<"j is greater";

This is C++, not C, try comp.lang.c++. Or use printf() or puts().
Also the else branch is wrong "j is greater or equal" would be the correct
statement.

> Since i is unsigned , j is greater . I know why , but vaguely . Can
> someone please explain the mechanics behind it . Is the unsigned int
> converted to signed or is it vice-versa . Also , the same rules would
> apply to char as well , right ?

in C, I think the signed int would get promoted to unsigned int. And yes,
chars should do the same. That's why you get 'j is greater' (in C, not sure
about C++)

if((int)i>j)
would give the correct/expected output.

Strange, I'd have expected my compiler to give a warning about the implicit
conversion ...


Bye, Jojo



Martin Wells 10-02-2007 09:46 PM

Re: Unsigned & signed int
 
On Oct 2, 7:34 pm, Kislay <kislaychan...@gmail.com> wrote:
> Consider the following code snippet
>
> unsigned int i=10;
> int j= - 2; // minus 2
> if(i>j)
> cout<<"i is greater";
> else
> cout<<"j is greater";
>
> Since i is unsigned , j is greater . I know why , but vaguely . Can
> someone please explain the mechanics behind it . Is the unsigned int
> converted to signed or is it vice-versa . Also , the same rules would
> apply to char as well , right ?



Rule 1: Things get promoted to SIGNED integer types where possible,
otherwise UNSIGNED.

Rule 2: The smaller of the two types involved in the operation has to
change to the bigger type.

Rule 3: If you're left with two types of the same size, but where one
is signed and the other unsigned, they both become unsigned.

Firstly, let's start with the types smaller than int. These types are
as follows:

signed char
unsigned char
signed short
unsigned short

The C Standard guarantees the following:
CHAR_MAX <= SHRT_MAX <= INT_MAX <= LONG_MAX
and also the following:
UCHAR_MAX <= USHRT_MAX <= UINT_MAX <= ULONG_MAX

Before you can do ANYTHING to any of the integer types smaller than
int, they must be promoted. As mentioned before, they promote to
SIGNED where possible, otherwise unsigned. Signed char and signed
short will always promote to signed int on every implementation. As
for unsigned char and unsigned short, they will promote to signed int
on some system, but unsigned int on other systems, depending on
whether all the values of the smaller type can be stored in signed
int. After the promotion of the smaller type takes place, you could
still be left with the following combinations:

1: (unsigned int) + (signed int)
2: (int) + (long)
3: (unsigned int) + (long)
4: (unsigned int) + (unsigned long)

In number 1, both types are the same size but one of them is unsigned,
so they both become unsigned.
In number 2, the int has to become a long.
In number 3, the int has to become a long, but we don't know if all
the values of unsigned int can be stored in a signed long. Therefore,
on some systems this will become an unsigned long, while on others it
will become a signed long. Once the two types are either kind of long,
they will become unsigned long if either of them is unsigned,
otherwise they'll stay as signed long.
In number 4, they'll both become unsigned long.

I suppose the thought process is as follows:

1: Promote things that are smaller than int.
2: Match the sizes (but being careful about whether the small type
will become signed or unsigned of the bigger type).
3: Once the sizes are match, match the signs (if one of them is
unsigned, then they both become unsigned).

Now... as for the rules for converting from unsigned to signed... what
you do is take the max value of the unsigned type, add 1 to it, and
then add it to the signed value.

So let's take -1. Let's pretend that UCHAR_MAX is 255. Therefore we
add as follows:

-1 + (255 + 1) = 255

Therefore the following two are equivalent:

char unsigned c = UCHAR_MAX;
char unsigned c = -1;

As are the following:

short unsigned su = USHRT_MAX - 5;
short unsigned su = -6;

I'll edit your own original code to show you exactly what's going to
happen:

unsigned int i=10;

int j= - 2; // minus 2

unsigned j_changed_to_unsigned = UINT_MAX - 1;

if(i > j_changed_to_unsigned)
cout<<"i is greater";
else
cout<<"j_changed_to_unsigned is greater";


You might wanna use printf instead of cout on this newsgroup though ;)

Martin



Martin Wells 10-02-2007 09:50 PM

Re: Unsigned & signed int
 

> Now... as for the rules for converting from unsigned to signed...



signed to unsigned


Martin Ambuhl 10-02-2007 10:39 PM

Re: Unsigned & signed int
 
Kislay wrote:
> Consider the following code snippet
>
> unsigned int i=10;
> int j= - 2; // minus 2
> if(i>j)
> cout<<"i is greater";
> else
> cout<<"j is greater";


Note that however cout is defined, and to use the "<<" operator on it is
really ought to be an unsigned integral type, it makes no sense to shift
it left by a string constant. And you probably want to lose the space
between '-' and '2', as well as the silly comment.

> Since i is unsigned , j is greater . I know why , but vaguely . Can
> someone please explain the mechanics behind it . Is the unsigned int
> converted to signed or is it vice-versa . Also , the same rules would
> apply to char as well , right ?


I believe the question you meant to ask is explicitly covered in the
FAQ. See Question 3.19 "Q: What's the difference between the ``unsigned
preserving'' and ``value preserving'' rules?" where, among other things,
this code
unsigned short us = 10;
int i = -5;
if(i > us)
printf("whoops!\n");
is discussed.
The question and answer can be found at
<http://c-faq.com/expr/preservingrules.html>, but you really want to
start trying to digest the FAQ as a whole <http://c-faq.com/index.html>.

pete 10-02-2007 11:05 PM

Re: Unsigned & signed int
 
Martin Wells wrote:

> The C Standard guarantees the following:
> CHAR_MAX <= SHRT_MAX <= INT_MAX <= LONG_MAX


You misspelled "SCHAR_MAX".

--
pete

Old Wolf 10-03-2007 04:00 AM

Re: Unsigned & signed int
 
On Oct 3, 10:46 am, Martin Wells <war...@eircom.net> wrote:
>
> char unsigned c = UCHAR_MAX;
> char unsigned c = -1;


JKop, JKop, is that you?



All times are GMT. The time now is 04:43 AM.

Powered by vBulletin®. Copyright ©2000 - 2014, vBulletin Solutions, Inc.
SEO by vBSEO ©2010, Crawlability, Inc.