about shifting

[lak]

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

 

[Pietro Cerutti]

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

 

[Richard Bos]

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

 

[Charlie Gordon]

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.

 

[Pietro Cerutti]

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

 

[Kenneth Brody]

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:[email protected]>

 

[CBFalconer]

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.

 

[Flash Gordon]

Pietro Cerutti wrote, On 20/09/07 14:45:

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]

Charlie Gordon wrote, On 20/09/07 13:51:

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.

 

[user923005]

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.

 

[Charlie Gordon]

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.

 

[Ark Khasin]

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

 

[Charlie Gordon]

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

 

[Ark Khasin]

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

 

[Army1987]

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?)

 

[Ben Bacarisse]

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.

 

[Tor Rustad]

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.