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hi
i have a problem with integer subtraction in C.
printf("%d", c < (a - b));
a is got from a #define and is 0x80000000 and b is got from input and
is also 0x80000000.
c is ffffffff (-1).
Now, this should print 1 (true) but it prints 0!

If I modify this to
d = c < (a - b);
printf("%d", d);
it prints 1 correctly.

I am stumped, please help! i guess something wrong happens during the
comparison, i dont know what.
If i just retain my second method which works, is everything ok? or is
there some risk?

thanks a lot
bruce

wrote:
> hi
> i have a problem with integer subtraction in C.
> printf("%d", c < (a - b));
> a is got from a #define and is 0x80000000 and b is got from input and
> is also 0x80000000.
> c is ffffffff (-1).
> Now, this should print 1 (true) but it prints 0!

You seem to have fallen into the pit too many C programmers
dig for themselves: Thinking about representations rather than
values. For example, `ffffffff' is *not* -1; it is (if there's
a `0x' at the beginning) 4294967295. Since `a-b' is zero and `c'
is this large positive number, the output is as expected.

> If I modify this to
> d = c < (a - b);
> printf("%d", d);
> it prints 1 correctly.

Did you make any other modifications? Wait: don't answer!
Instead, please post an *actual* *complete* *minimal* example
of the code that baffles you. I'm trying to guess what the
#define's and variable declarations and initializations and
assignments actually looked like, and if my guesses are wide of
the mark I'll give a diagnosis that has nothing to do with your
actual problem. Show the actual problem and somebody will be
likely to help.

Consider: If you're baffled by the inexplicable behavior
of some piece of code, are you in a good position to winnow the
relevant from the irrelevant pieces? No? Then don't try; you're
quite likely (as in this instance) to leave out the important bits.

> I am stumped, please help! i guess something wrong happens during the
> comparison, i dont know what.
> If i just retain my second method which works, is everything ok? or is
> there some risk?

Impossible to say on the scanty evidence available.

--
Eric Sosman
lid

wrote:
> hi
> i have a problem with integer subtraction in C.
> printf("%d", c < (a - b));
> a is got from a #define and is 0x80000000 and b is got from input and
> is also 0x80000000.
> c is ffffffff (-1).
> Now, this should print 1 (true) but it prints 0!

Are you _sure_ that an int can hold such values?

The following code should produce, on most implementations, a number of
warnings. Pay attention to the warnings, and then think about what they
and the possibly strange-looking output might mean:

#include <stdio.h>

int main(void)
{
unsigned short usc = 0xffffffff, usa = 0x80000000, usb = 0x80000000;
/* the above line leads to truncation in my implementation, shorts
not being 32 bits */
unsigned long ulc = 0xffffffffL, ula = 0x80000000L, ulb =
0x80000000L;
signed short ssc = 0xffffffff, ssa = 0x80000000, ssb = 0x80000000;
/* the above line leads to truncation in my implementation, shorts
not being 32 bits */
signed long slc = 0xffffffffL, sla = 0x80000000L, slb = 0x80000000L;
printf("[Output for this implementation]\n");
printf("%d < (%d - %d)? %s\n",
ssc, ssa, ssb, (ssc < (ssa - ssb)) ? "yes" : "no");
printf("%ld < (%ld - %ld)? %s\n",
slc, sla, slb, (slc < (sla - slb)) ? "yes" : "no");
printf("%u < (%u - %u) or %x < (%x - %x)? %s\n",
usc, usa, usb,
usc, usa, usb, (usc < (usa - usb)) ? "yes" : "no");
printf("%lu < (%lu - %lu) or %lx < (%lx - %lx)? %s\n\n",
ulc, ula, ulb,
ulc, ula, ulb, (ulc < (ula - ulb)) ? "yes" : "no");

usc = 0xffff;
usa = 0x8000;
usb = 0x8000;
ulc = 0xffffL;
ula = 0x8000L;
ulb = 0x8000L;
ssc = 0xffff;
ssa = 0x8000;
ssb = 0x8000;
/* the above lines lead to truncation in my implementation, shorts
not being 32 bits */
slc = 0xffffL;
sla = 0x8000L;
slb = 0x8000L;
printf("%d < (%d - %d)? %s\n",
ssc, ssa, ssb, (ssc < (ssa - ssb)) ? "yes" : "no");
printf("%ld < (%ld - %ld)? %s\n",
slc, sla, slb, (slc < (sla - slb)) ? "yes" : "no");
printf("%u < (%u - %u) or %x < (%x - %x)? %s\n",
usc, usa, usb,
usc, usa, usb, (usc < (usa - usb)) ? "yes" : "no");
printf("%lu < (%lu - %lu) or %lx < (%lx - %lx)? %s\n\n",
ulc, ula, ulb,
ulc, ula, ulb, (ulc < (ula - ulb)) ? "yes" : "no");
return 0;
}


[Output for this implementation]
-1 < (0 - 0)? yes
-1 < (-2147483648 - -214748364? yes
65535 < (0 - 0) or ffff < (0 - 0)? no
4294967295 < (2147483648 - 214748364 or ffffffff < (80000000 -
80000000)? no

-1 < (-32768 - -3276? yes
65535 < (32768 - 3276? no
65535 < (32768 - 3276 or ffff < (8000 - 8000)? no
65535 < (32768 - 3276 or ffff < (8000 - 8000)? no

Eric,
Thanks a lot! I solved the problem but I still have a question.
>For example, `ffffffff' is *not* -1; it is (if there's a `0x' at the
beginning)
> 4294967295.

The problem was this:
As mentioned before, I was getting the value of 'a' from a #define. (c
and b are ok since they are got from input and are signed numbers)

#define a 0x80000000
assigned not -2147483648 but some other big positive value to 'a'.

Now, I removed the #define and put this as a statement inside my
function and it works.

But lint gives a warning saying that "initializer is out of range:
0x80000000"

How can I fix it?

Thanks
Bruce

Thanks a lot! I solved the problem but I still have a question.

>For example, `ffffffff' is *not* -1; it is (if there's a `0x' at the
beginning)
> 4294967295.

The problem was this:
As mentioned before, I was getting the value of 'a' from a #define. (c
and b are ok since they are got from input and are signed numbers)

#define a 0x80000000
assigned not -2147483648 but some other big positive value to 'a'.

Now, I removed the #define and put this as a statement inside my
function and it works.

But lint gives a warning saying that "initializer is out of range:
0x80000000"

How can I fix it?

Thanks
Bruce

Thanks for the hints. I solved the problem but now I have another
problem!
In
printf("%d", c < (a - b));
'a' is got from #define as I mentioned earlier, now all that is in the
#define is treated as unsigned! So
#define a 0x800000000
assigns not -2147483648 but some big positive value. And this caused
problems in my calculation with signed numbers.
I fixed this by saying
int a = -2147483648 ; inside my function.
but C gives a warningsaying it is out of range. I searched for this new
problem, know why it occurs, but what would be the best solution?

wrote:
> Eric,
> Thanks a lot! I solved the problem but I still have a question. [...]

... and you have failed to pay any heed to a crucial
part of my earlier response, to wit

>> [...] please post an *actual* *complete* *minimal* example
>> of the code that baffles you. [...]

> How can I fix it?

My crystal ball says your error is on line 42.

--
Eric Sosman
lid

Eric,
The reason I did not heed your suggestion is simple, the problem got
solved but turned into another. which is just

"int a = 0x80000000;"
gives a warning since a is -2147483648 which is split as an operator
and a positive number by C.

How to remove this warning? (a = -2147483647 - 1 looks dirty)


Eric Sosman wrote:
> wrote:
> > Eric,
> > Thanks a lot! I solved the problem but I still have a question.
[...]
>
> ... and you have failed to pay any heed to a crucial
> part of my earlier response, to wit
>
> >> [...] please post an *actual* *complete* *minimal* example
> >> of the code that baffles you. [...]
>
> > How can I fix it?
>
> My crystal ball says your error is on line 42.
>
> --
> Eric Sosman
> lid

wrote in news:1112567684.942695.52570
@f14g2000cwb.googlegroups.com:

Eric Sosman wrote:

>> >> [...] please post an *actual* *complete* *minimal* example
>> >> of the code that baffles you. [...]
....
>> ... and you have failed to pay any heed to a crucial
>> part of my earlier response, to wit

> Eric,
> The reason I did not heed your suggestion is simple, the problem got
> solved but turned into another. which is just

How does that eliminate the importance of posting that satisfies the the
three important criteria listed above???

> "int a = 0x80000000;"

#include <stdio.h>
#include <limits.h>

int main(void) {
printf("The maximum possible value for an int is: %d\n", INT_MAX);
printf("The minimum possible value for an int is: %d\n", INT_MIN);
return 0;
}

The value you are assigning to a, 0x80000000 is an unsigned literal
whose value does not fit in a signed int.

If you want to assign -1 to a, then you should do exactly that:

int a = -1;

Now, that is not so hard, is it?

Sinan
--
A. Sinan Unur <>
(reverse each component and remove .invalid for email address)

comp.lang.perl.misc guidelines on the WWW:
http://mail.augustmail.com/~tadmc/cl...uidelines.html

Sinan,
I dont think u got the problem. The value I want in 'a' is -2147483648,
not -1.

int a = -2147483648;
will generate a warning.

int a = 0x80000000;
will also generate a warning.

how to eliminate the warning, is my query.

A. Sinan Unur wrote:
> wrote in news:1112567684.942695.52570
> @f14g2000cwb.googlegroups.com:
>
> Eric Sosman wrote:
>
> >> >> [...] please post an *actual* *complete* *minimal* example
> >> >> of the code that baffles you. [...]
> ...
> >> ... and you have failed to pay any heed to a crucial
> >> part of my earlier response, to wit
>
> > Eric,
> > The reason I did not heed your suggestion is simple, the problem
got
> > solved but turned into another. which is just
>
> How does that eliminate the importance of posting that satisfies the
the
> three important criteria listed above???
>
> > "int a = 0x80000000;"
>
> #include <stdio.h>
> #include <limits.h>
>
> int main(void) {
> printf("The maximum possible value for an int is: %d\n",
INT_MAX);
> printf("The minimum possible value for an int is: %d\n",
INT_MIN);
> return 0;
> }
>
> The value you are assigning to a, 0x80000000 is an unsigned literal
> whose value does not fit in a signed int.
>
> If you want to assign -1 to a, then you should do exactly that:
>
> int a = -1;
>
> Now, that is not so hard, is it?
>
> Sinan
> --
> A. Sinan Unur <>
> (reverse each component and remove .invalid for email address)
>
> comp.lang.perl.misc guidelines on the WWW:
> http://mail.augustmail.com/~tadmc/cl...uidelines.html