Discussion in 'C Programming' started by lak, Sep 20, 2007.

1. ### lakGuest

i know left and right shift normally,but i cant know what happens if
it is negative.
for example
int x=-2;
x<<=1;//what happens here

lak, Sep 20, 2007

2. ### Pietro CeruttiGuest

lak wrote:
> i know left and right shift normally,but i cant know what happens if
> it is negative.
> for example
> int x=-2;
> x<<=1;//what happens here
>

nothing different from usual.. the bit representation of -2 is shifted
one bit to the left...

> cat test_shift.c

#include <stdio.h>
int main(void)
{

int k = -1;
printf("k is %d (%x)\n", k, k);
k<<=4;
printf("k is %d (%x)\n", k, k);

return (0);
}

> gcc -Wall -o test_shift test_shift.c && ./test_shift

k is -1 (ffffffff)
k is -16 (fffffff0)

Pietro Cerutti

Pietro Cerutti, Sep 20, 2007

3. ### Richard BosGuest

lak <> wrote:

> i know left and right shift normally,but i cant know what happens if
> it is negative.
> for example
> int x=-2;
> x<<=1;//what happens here

That is correct: you cannot know that.

From paragraph 6.5.7 in the ISO C Standard:

# 4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated
# bits are filled with zeros. If E1 has an unsigned type, the value of
# the result is E1 x 2 E2 ,reduced modulo one more than the maximum
# value representable in the result type. If E1 has a signed type and
# nonnegative value, and E1 x 2 E2 is representable in the result
# type, then that is the resulting value; otherwise, the behavior is
# undefined. ^^^^^^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^

Note the under^^^lined bit. Since in your case x is neither an unsigned
integer, nor a signed integer with a positive value, the behaviour of
your code is undefined; and this means that, as far as ISO C is
concerned, you cannot know what happens. (It may be possible to discover
what happens on a particular computer using a particular compiler with
particular compilation settings, but I advise against it; on the next
system, or even on the next level of optimisation, the result can easily
be different.)

Richard

Richard Bos, Sep 20, 2007
4. ### Charlie GordonGuest

"Richard Bos" <> a écrit dans le message de news:
4all.nl...
> lak <> wrote:
>
>> i know left and right shift normally,but i cant know what happens if
>> it is negative.
>> for example
>> int x=-2;
>> x<<=1;//what happens here

>
> That is correct: you cannot know that.
>
> From paragraph 6.5.7 in the ISO C Standard:
>
> # 4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated
> # bits are filled with zeros. If E1 has an unsigned type, the value of
> # the result is E1 x 2 E2 ,reduced modulo one more than the maximum
> # value representable in the result type. If E1 has a signed type and
> # nonnegative value, and E1 x 2 E2 is representable in the result
> # type, then that is the resulting value; otherwise, the behavior is
> # undefined. ^^^^^^^^^^^^^^^^^^^^^^^^^^
> ^^^^^^^^^
>
> Note the under^^^lined bit. Since in your case x is neither an unsigned
> integer, nor a signed integer with a positive value, the behaviour of
> your code is undefined; and this means that, as far as ISO C is
> concerned, you cannot know what happens. (It may be possible to discover
> what happens on a particular computer using a particular compiler with
> particular compilation settings, but I advise against it; on the next
> system, or even on the next level of optimisation, the result can easily
> be different.)

Richard is correct.
However, if you expected x <<= 1 to be equivalent to x += x, as would
"normally" be the case on regular 2s-complement machines, you might as well
write the latter.

--
Chqrlie.

Charlie Gordon, Sep 20, 2007
5. ### Pietro CeruttiGuest

Pietro Cerutti wrote:
> lak wrote:
>> i know left and right shift normally,but i cant know what happens if
>> it is negative.
>> for example
>> int x=-2;
>> x<<=1;//what happens here
>>

>
> nothing different from usual.. the bit representation of -2 is shifted
> one bit to the left...

Umh, I have to apologize.. my sentence is actually incorrect. That's
true for right-shifting, while for left-shifting a negative left-hand
operand invokes UB

--
Pietro Cerutti

Pietro Cerutti, Sep 20, 2007
6. ### Kenneth BrodyGuest

Nit-pick non-UB (was Re: about shifting)

Charlie Gordon wrote:
[... bit-shifting negative numbers ...]
> However, if you expected x <<= 1 to be equivalent to x += x, as would
> "normally" be the case on regular 2s-complement machines, you might
> as well write the latter.

Okay, nit-pick time related to UB.

Why doesn't the statement:

x += x;

violate 6.5p2:

Between the previous and next sequence point an object shall
have its stored value modified at most once by the evaluation
of an expression. Furthermore, the prior value shall be read
only to determine the value to be stored.

Yes, I see that footnote 71 says that "i = i + 1" is allowed by the
paragraph, but why is the "x" on the right of "+=" not violating the
"shall be read only to determine the value to be stored"? How is
this different from "y = x + x++;" in the use of "x"?

Obviously, something like "x += x;" must be allowed, but what is it
about 6.5p2 that allows it?

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer.h> |
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:>

Kenneth Brody, Sep 20, 2007
7. ### CBFalconerGuest

Pietro Cerutti wrote:
> Pietro Cerutti wrote:
>> lak wrote:
>>
>>> i know left and right shift normally,but i cant know what
>>> happens if it is negative. for example
>>> int x=-2;
>>> x<<=1;//what happens here

>>
>> nothing different from usual.. the bit representation of -2 is
>> shifted one bit to the left...

>
> Umh, I have to apologize.. my sentence is actually incorrect.
> That's true for right-shifting, while for left-shifting a
> negative left-hand operand invokes UB

Since lak appears to be a newbie, explain that UB means "undefined
behaviour". In other words, don't do that. Also for positive
operands that overflow.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

--
Posted via a free Usenet account from http://www.teranews.com

CBFalconer, Sep 20, 2007
8. ### Flash GordonGuest

Pietro Cerutti wrote, On 20/09/07 14:45:
> Pietro Cerutti wrote:
>> lak wrote:
>>> i know left and right shift normally,but i cant know what happens if
>>> it is negative.
>>> for example
>>> int x=-2;
>>> x<<=1;//what happens here
>>>

>> nothing different from usual.. the bit representation of -2 is shifted
>> one bit to the left...

>
> Umh, I have to apologize.. my sentence is actually incorrect. That's
> true for right-shifting, while for left-shifting a negative left-hand
> operand invokes UB

For right shifting it is implementation defined, so you were just wrong.
--
Flash Gordon

Flash Gordon, Sep 20, 2007
9. ### Flash GordonGuest

Charlie Gordon wrote, On 20/09/07 13:51:
> "Richard Bos" <> a écrit dans le message de news:
> 4all.nl...
>> lak <> wrote:
>>
>>> i know left and right shift normally,but i cant know what happens if
>>> it is negative.
>>> for example
>>> int x=-2;
>>> x<<=1;//what happens here

>> That is correct: you cannot know that.

<snip>

> Richard is correct.
> However, if you expected x <<= 1 to be equivalent to x += x, as would
> "normally" be the case on regular 2s-complement machines, you might as well
> write the latter.

Personally I would write it as x *= 2.
--
Flash Gordon

Flash Gordon, Sep 20, 2007
10. ### user923005Guest

Re: Nit-pick non-UB (was Re: about shifting)

On Sep 20, 10:33 am, Kenneth Brody <> wrote:
> Charlie Gordon wrote:
>
> [... bit-shifting negative numbers ...]
>
> > However, if you expected x <<= 1 to be equivalent to x += x, as would
> > "normally" be the case on regular 2s-complement machines, you might
> > as well write the latter.

>
> Okay, nit-pick time related to UB.
>
> Why doesn't the statement:
>
> x += x;
>
> violate 6.5p2:
>
> Between the previous and next sequence point an object shall
> have its stored value modified at most once by the evaluation
> of an expression. Furthermore, the prior value shall be read
> only to determine the value to be stored.
>
> Yes, I see that footnote 71 says that "i = i + 1" is allowed by the
> paragraph, but why is the "x" on the right of "+=" not violating the
> "shall be read only to determine the value to be stored"? How is
> this different from "y = x + x++;" in the use of "x"?
>
> Obviously, something like "x += x;" must be allowed, but what is it
> about 6.5p2 that allows it?

"Furthermore, the prior value shall be read only to determine the
value to be stored."

Quite frankly, in this case, I simply don't see how it would be
possible for the compiler to get it wrong.

For an instance like:

i = ++i;

there are two modifications of i, so it's right out.

But how is:
x += x;
more dangerous than (for instance):
x = x;
Both of which have to examine the contents of x use those contents to
modify x (though this second instance can be thrown out by the
compiler if it chooses because of the 'as if' rule.)

For the instance of:
y = x + x++;
We don't even need the y. This is also undefined behavior:

#include <stdlib.h>
int t(void)
{
int x = rand();
return x + x++;
}

We are adding x + <something>
We are also incrementing x.
There is no sequence point.

user923005, Sep 20, 2007
11. ### Charlie GordonGuest

Re: Nit-pick non-UB (was Re: about shifting)

"Kenneth Brody" <> a écrit dans le message de news:
...
> Charlie Gordon wrote:
> [... bit-shifting negative numbers ...]
>> However, if you expected x <<= 1 to be equivalent to x += x, as would
>> "normally" be the case on regular 2s-complement machines, you might
>> as well write the latter.

>
> Okay, nit-pick time related to UB.
>
> Why doesn't the statement:
>
> x += x;
>
> violate 6.5p2:
>
> Between the previous and next sequence point an object shall
> have its stored value modified at most once by the evaluation
> of an expression. Furthermore, the prior value shall be read
> only to determine the value to be stored.
>
> Yes, I see that footnote 71 says that "i = i + 1" is allowed by the
> paragraph, but why is the "x" on the right of "+=" not violating the
> "shall be read only to determine the value to be stored"? How is
> this different from "y = x + x++;" in the use of "x"?
>
> Obviously, something like "x += x;" must be allowed, but what is it
> about 6.5p2 that allows it?

x is modified only once, and its value is read only to determine the value
to be stored, once or twice depending on quality of implementation or
presence of volatile qualifier on x's definition.

Change x <<= N into x *= 1 << N to get rid of the problem with negative x,
as long as the multiplication does not overflow. Also note that *all*
current architectures use two's complement representation for integers, and
implement left shifting on negative numbers consistently.

--
Chqrlie.

Charlie Gordon, Sep 20, 2007
12. ### Ark KhasinGuest

Richard Bos wrote:
> lak <> wrote:
>
>> i know left and right shift normally,but i cant know what happens if
>> it is negative.
>> for example
>> int x=-2;
>> x<<=1;//what happens here

>
> That is correct: you cannot know that.
>
> From paragraph 6.5.7 in the ISO C Standard:
>
> # 4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated
> # bits are filled with zeros. If E1 has an unsigned type, the value of
> # the result is E1 x 2 E2 ,reduced modulo one more than the maximum
> # value representable in the result type. If E1 has a signed type and
> # nonnegative value, and E1 x 2 E2 is representable in the result
> # type, then that is the resulting value; otherwise, the behavior is
> # undefined. ^^^^^^^^^^^^^^^^^^^^^^^^^^
> ^^^^^^^^^
>
> Note the under^^^lined bit. Since in your case x is neither an unsigned
> integer, nor a signed integer with a positive value, the behaviour of
> your code is undefined; and this means that, as far as ISO C is
> concerned, you cannot know what happens. (It may be possible to discover
> what happens on a particular computer using a particular compiler with
> particular compilation settings, but I advise against it; on the next
> system, or even on the next level of optimisation, the result can easily
> be different.)
>
> Richard

I am afraid I don't understand shifts anymore. I suppose(d) that on
32-bit machine
insigned x, y;
y = 32U; //where the compiler doesn't see it
(x<<y)==0U && (x>>y)==0U
All ARM compilers I've seen, the effective calculation is
x << (y & 31U) and x << (y & 31U)
(because that's how the CPU instructions work)
Is this compliant?

-- Ark

Ark Khasin, Sep 21, 2007
13. ### Charlie GordonGuest

"Ark Khasin" <> a écrit dans le message de news:
4sEIi.6556\$Yb2.2730@trndny08...
> Richard Bos wrote:
>> lak <> wrote:
>>
>>> i know left and right shift normally,but i cant know what happens if
>>> it is negative.
>>> for example
>>> int x=-2;
>>> x<<=1;//what happens here

>>
>> That is correct: you cannot know that. From paragraph 6.5.7 in the ISO C
>> Standard:
>>
>> # 4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated
>> # bits are filled with zeros. If E1 has an unsigned type, the value of
>> # the result is E1 x 2 E2 ,reduced modulo one more than the maximum
>> # value representable in the result type. If E1 has a signed type and
>> # nonnegative value, and E1 x 2 E2 is representable in the result
>> # type, then that is the resulting value; otherwise, the behavior is
>> # undefined. ^^^^^^^^^^^^^^^^^^^^^^^^^^
>> ^^^^^^^^^
>>
>> Note the under^^^lined bit. Since in your case x is neither an unsigned
>> integer, nor a signed integer with a positive value, the behaviour of
>> your code is undefined; and this means that, as far as ISO C is
>> concerned, you cannot know what happens. (It may be possible to discover
>> what happens on a particular computer using a particular compiler with
>> particular compilation settings, but I advise against it; on the next
>> system, or even on the next level of optimisation, the result can easily
>> be different.)
>>
>> Richard

> I am afraid I don't understand shifts anymore. I suppose(d) that on 32-bit
> machine
> insigned x, y;
> y = 32U; //where the compiler doesn't see it
> (x<<y)==0U && (x>>y)==0U
> All ARM compilers I've seen, the effective calculation is
> x << (y & 31U) and x << (y & 31U)
> (because that's how the CPU instructions work)
>
> Is this compliant?

C99 6.5.7p3 also says: If the value of the right operand is negative or is
greater than or equal to the width of the promoted left operand, the
behavior is undefined. So shifting a 32-bit integer by 32 invokes undefined
behaviour. That covers any harware behaviour whatsoever.

If you really want that, you need to split the operation:
(x << 16) << 16 == 0 && (x >> 16) >> 16 == 0

--
Chqrlie.

Charlie Gordon, Sep 21, 2007
14. ### Ark KhasinGuest

Charlie Gordon wrote:
> "Ark Khasin" <> a écrit dans le message de news:
> 4sEIi.6556\$Yb2.2730@trndny08...
>> Richard Bos wrote:
>>> lak <> wrote:
>>>
>>>> i know left and right shift normally,but i cant know what happens if
>>>> it is negative.
>>>> for example
>>>> int x=-2;
>>>> x<<=1;//what happens here
>>> That is correct: you cannot know that. From paragraph 6.5.7 in the ISO C
>>> Standard:
>>>
>>> # 4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated
>>> # bits are filled with zeros. If E1 has an unsigned type, the value of
>>> # the result is E1 x 2 E2 ,reduced modulo one more than the maximum
>>> # value representable in the result type. If E1 has a signed type and
>>> # nonnegative value, and E1 x 2 E2 is representable in the result
>>> # type, then that is the resulting value; otherwise, the behavior is
>>> # undefined. ^^^^^^^^^^^^^^^^^^^^^^^^^^
>>> ^^^^^^^^^
>>>
>>> Note the under^^^lined bit. Since in your case x is neither an unsigned
>>> integer, nor a signed integer with a positive value, the behaviour of
>>> your code is undefined; and this means that, as far as ISO C is
>>> concerned, you cannot know what happens. (It may be possible to discover
>>> what happens on a particular computer using a particular compiler with
>>> particular compilation settings, but I advise against it; on the next
>>> system, or even on the next level of optimisation, the result can easily
>>> be different.)
>>>
>>> Richard

>> I am afraid I don't understand shifts anymore. I suppose(d) that on 32-bit
>> machine
>> insigned x, y;
>> y = 32U; //where the compiler doesn't see it
>> (x<<y)==0U && (x>>y)==0U
>> All ARM compilers I've seen, the effective calculation is
>> x << (y & 31U) and x << (y & 31U)
>> (because that's how the CPU instructions work)
>>
>> Is this compliant?

>
> C99 6.5.7p3 also says: If the value of the right operand is negative or is
> greater than or equal to the width of the promoted left operand, the
> behavior is undefined. So shifting a 32-bit integer by 32 invokes undefined
> behaviour. That covers any harware behaviour whatsoever.
>
> If you really want that, you need to split the operation:
> (x << 16) << 16 == 0 && (x >> 16) >> 16 == 0
>

Thanks. I should have known better indeed.
-- Ark

Ark Khasin, Sep 21, 2007
15. ### Army1987Guest

Re: Nit-pick non-UB (was Re: about shifting)

On Thu, 20 Sep 2007 13:33:21 -0400, Kenneth Brody wrote:

> Why doesn't the statement:
>
> x += x;
>
> violate 6.5p2:
>
> Between the previous and next sequence point an object shall
> have its stored value modified at most once by the evaluation
> of an expression. Furthermore, the prior value shall be read
> only to determine the value to be stored.
>
> Yes, I see that footnote 71 says that "i = i + 1" is allowed by the
> paragraph, but why is the "x" on the right of "+=" not violating the
> "shall be read only to determine the value to be stored"? How is
> this different from "y = x + x++;" in the use of "x"?
>
> Obviously, something like "x += x;" must be allowed, but what is it
> about 6.5p2 that allows it?

x is modified only once, and it is only read to determine the
value to be stored. Right? I think the last sentence in that quote
is very mysterious. There have been endless discussions about
whether list = list->next = malloc(sizeof *list) causes UB. And a
sufficiently bizarre interpretation of that sentence would mean
that
x = 0 * x + rand() causes UB, and
x = 1 * x + abs(0) doesn't. I definitely think that this isn't the
intent. Anyway, I've developed a paranoia of not using the same
lvalue twice in an expression, e.g. I'd rather write i *= -1 than
i = -i, or u ^= UINT_MAX than u = ~u. (At least this helps to
write macros which evaluate its argument exactly once, but is it
*really* necessary when written directly in the code?)

--
Army1987 (Replace "NOSPAM" with "email")
If you're sending e-mail from a Windows machine, turn off Microsoft's
stupid â€œSmart Quotesâ€ feature. This is so you'll avoid sprinkling garbage
characters through your mail. -- Eric S. Raymond and Rick Moen

Army1987, Sep 21, 2007
16. ### Ben BacarisseGuest

Re: Nit-pick non-UB (was Re: about shifting)

Kenneth Brody <> writes:

> Charlie Gordon wrote:
> [... bit-shifting negative numbers ...]
>> However, if you expected x <<= 1 to be equivalent to x += x, as would
>> "normally" be the case on regular 2s-complement machines, you might
>> as well write the latter.

>
> Okay, nit-pick time related to UB.
>
> Why doesn't the statement:
>
> x += x;
>
> violate 6.5p2:
>
> Between the previous and next sequence point an object shall
> have its stored value modified at most once by the evaluation
> of an expression. Furthermore, the prior value shall be read
> only to determine the value to be stored.
>
> Yes, I see that footnote 71 says that "i = i + 1" is allowed by the
> paragraph, but why is the "x" on the right of "+=" not violating the
> "shall be read only to determine the value to be stored"? How is
> this different from "y = x + x++;" in the use of "x"?

Why 'x += x;' is OK has been answered, but why 'y = x + x++;' is not
might need a bit more discussion. Here, the prior value is read to
determine both the value to be stored in x and the value of x for the
addition. The term "read" is important. If the standard had said
"used" then 'y = 1 + x++;' would be UB (x's prior value is used for a
purpose other than to determine the value to be stored).

The prohibition is on more than one part of the expression referring
to the value of an object whose value the expression modifies. I
doubt that that wording is any clearer (or I am sure the committee
would have used it) but it gives another way to look at it.

--
Ben.

Ben Bacarisse, Sep 21, 2007
17. ### Tor RustadGuest

lak wrote:
> i know left and right shift normally,but i cant know what happens if
> it is negative.
> for example
> int x=-2;
> x<<=1;//what happens here

A simple rule of thumb, is to do bitwise operations on unsigned types.

--
Tor <torust [at] online [dot] no>

Tor Rustad, Sep 21, 2007