Rounding during division

[Christopher Key]

Hello,

Could anyone tell me what the various C standards say on the subject of
rounding during signed integer division. On my system, it always rounds
towards 0:

printf("%d", 3/2); // 1
printf("%d", 1/2); // 0
printf("%d", -1/2); // 0
printf("%d", -3/2); // -1

Can I rely on this always being the case?
Secondly, is there anything that I can rely when right shifting a
negative int32_t. Given that int32_t is guaranteed to be stored in
two's complement format, can it be assumed that right shifting by N is
the same as division by 2^N with rounding towards -inf? e.g:

printf("%d", 3>>1); // 1
printf("%d", 1>>1); // 0
printf("%d", -1>>1); // -1
printf("%d", -3>>1); // -3

If bitwise operations have undefined behaviour on negative intN_t types,
what was the point in specifying that they must be in two's complement
format?

Regards,

Chris

[Army1987]

On Wed, 12 Sep 2007 13:39:26 +0100, Christopher Key wrote:

Hello,

Could anyone tell me what the various C standards say on the subject of
rounding during signed integer division. On my system, it always rounds
towards 0:

printf("%d", 3/2); // 1
printf("%d", 1/2); // 0
printf("%d", -1/2); // 0
printf("%d", -3/2); // -1

Can I rely on this always being the case?

Yes.

Secondly, is there anything that I can rely when right shifting a
negative int32_t. Given that int32_t is guaranteed to be stored in
two's complement format, can it be assumed that right shifting by N is
the same as division by 2^N with rounding towards -inf? e.g:

printf("%d", 3>>1); // 1
printf("%d", 1>>1); // 0
printf("%d", -1>>1); // -1
printf("%d", -3>>1); // -3

If bitwise operations have undefined behaviour on negative intN_t types,
what was the point in specifying that they must be in two's complement
format?

Bitwise shifts on signed integers are implementation defined. For
example the sign bit may be treated differently.
--
Army1987 (Replace "NOSPAM" with "email")
If you're sending e-mail from a Windows machine, turn off Microsoft's
stupid “Smart Quotes� feature. This is so you'll avoid sprinkling garbage
characters through your mail. -- Eric S. Raymond and Rick Moen

[Ralf Damaschke]

Army1987 wrote:

On Wed, 12 Sep 2007 13:39:26 +0100, Christopher Key wrote:

>Hello,

Could anyone tell me what the various C standards say on the
subject of rounding during signed integer division. On my
system, it always rounds towards 0:

printf("%d", 3/2); // 1
printf("%d", 1/2); // 0
printf("%d", -1/2); // 0
printf("%d", -3/2); // -1

Can I rely on this always being the case?

Yes.

No. That is only true for C99.
There are still many C89-C95 implementations and they were
allowed to make their own decision:

| If either operand is negative, whether the
| result of the / operator is the largest integer less than the
| algebraic quotient or the smallest integer greater than the
| algebraic quotient is implementation-defined, as is the sign
| of the result of the % operator.

Ralf

[Mark McIntyre]

On Wed, 12 Sep 2007 15:13:04 +0200, in comp.lang.c , Army1987
<ar******@NOSPAM.itwrote:

>On Wed, 12 Sep 2007 13:39:26 +0100, Christopher Key wrote:

>Hello,

Could anyone tell me what the various C standards say on the subject of
rounding during signed integer division. On my system, it always rounds
towards 0:

printf("%d", 3/2); // 1
printf("%d", 1/2); // 0
printf("%d", -1/2); // 0
printf("%d", -3/2); // -1

Can I rely on this always being the case?

Yes.

No. For the positive cases, you can rely on it. For the negative
cases, compilers may round up or down, depending on which version of
the C standard they comply to.
--
Mark McIntyre

"Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it."
--Brian Kernighan

[Old Wolf]

On Sep 13, 1:13 am, Army1987 <[e-mail]:

On Wed, 12 Sep 2007 13:39:26 +0100, Christopher Key wrote:

Could anyone tell me what the various C standards say on the subject of
rounding during signed integer division. On my system, it always rounds
towards 0:

Can I rely on this always being the case?

Yes.

In fact it is implementation-defined in C90, whether it
rounds up or down for negative intergers.

Bitwise shifts on signed integers are implementation defined.

No. There are four cases:
1. Right-shift of non-negative value: well-defined
2. Right-shift of negative value: implementation-defined
3. Left-shift of negative value: undefined
4. Left-shift of non-negative value: undefined if the
shift would cause an overflow, otherwise well-defined

[Justin Spahr-Summers]

On Sep 17, 4:18 am, Christopher Key <[e-mail]:

Is there any clean, portable way of acheiving signed integer division by
a power of 2 with rounding towards -inf (under C99). gcc guarantees
that right-shifts on a signed integer will always give this behaviour,
and I'm guessing that pretty much any code compiled for a target with an
ASR instruction will do the same, but I really don't want the code to
silently break for compilers that don't.

The options seem to be:

1) Discover how right shift of negative values is implemented using
preprocessor macros and issue an error at compile time if required. I
don't think this is possible, although a useful extension to the C
standard might be to force the provision of macros that allow code to
determine how implementation defined behaviour has been implemented.

typedef int assertion_dummy_array[(-value >shiftbits == othervalue)
* 2 - 1];

If the condition given in the parentheses evaluates to false (0), then
the array typedef will have a negative size, which should issue a
compiler diagnostic.

[Christopher Key]

Justin Spahr-Summers wrote:

On Sep 17, 4:18 am, Christopher Key <[e-mail]:

>Is there any clean, portable way of acheiving signed integer division by
a power of 2 with rounding towards -inf (under C99). gcc guarantees
that right-shifts on a signed integer will always give this behaviour,
and I'm guessing that pretty much any code compiled for a target with an
ASR instruction will do the same, but I really don't want the code to
silently break for compilers that don't.

The options seem to be:

1) Discover how right shift of negative values is implemented using
preprocessor macros and issue an error at compile time if required. I
don't think this is possible, although a useful extension to the C
standard might be to force the provision of macros that allow code to
determine how implementation defined behaviour has been implemented.

typedef int assertion_dummy_array[(-value >shiftbits == othervalue)
* 2 - 1];

If the condition given in the parentheses evaluates to false (0), then
the array typedef will have a negative size, which should issue a
compiler diagnostic.

I'll give that a go. It'd be interesting to know how many compilers
warn correctly in all cases, even when cross compiling for a platform
with different shift instructions available to those on the host.

Regards,

Chris