# The strangest problem....

Discussion in 'C Programming' started by Jannick, Oct 9, 2003.

1. ### JannickGuest

Hi I have:
unsigned int myvalue=0;
unsigned char mytest=0;

mytest=0x34;
myvalue = mytest<<24;

Then myvalue gets the value 0x1A000000!!!!!
How is this possible? Should'nt it be 0x34000000???

Best Regards
Jannick

Jannick, Oct 9, 2003

2. ### Allan BruceGuest

"Jannick" <> wrote in message
news:bm3cko\$qan\$-c.dk...
> Hi I have:
> unsigned int myvalue=0;
> unsigned char mytest=0;
>
> mytest=0x34;
> myvalue = mytest<<24;
>
> Then myvalue gets the value 0x1A000000!!!!!
> How is this possible? Should'nt it be 0x34000000???
>
> Best Regards
> Jannick
>
>

Are you sure, its not a typo in your program? 0x1A000000 is 0x34<<23
Allan

Allan Bruce, Oct 9, 2003

3. ### Robert StankowicGuest

"Jannick" <> schrieb im Newsbeitrag
news:bm3cko\$qan\$-c.dk...
> Hi I have:
> unsigned int myvalue=0;
> unsigned char mytest=0;
>
> mytest=0x34;
> myvalue = mytest<<24;
>
> Then myvalue gets the value 0x1A000000!!!!!
> How is this possible? Should'nt it be 0x34000000???

Yes, it should (and is on my system, which of course doesn't proof
anything).

from the 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."

The only explanation I have - besides a typo (23 instead of 24) in your
actual code is something else in the code causing UB. I can't imagine, that
your compiler is broken in _such_ a way.
Can you post a minimum _compilable_ program which exhibits the error?

Robert

Robert Stankowic, Oct 9, 2003
4. ### hongkyGuest

my VC+win32 is right,
myvalue=0x34

"Jannick" <> wrote in message
news:bm3cko\$qan\$-c.dk...
> Hi I have:
> unsigned int myvalue=0;
> unsigned char mytest=0;
>
> mytest=0x34;
> myvalue = mytest<<24;
>
> Then myvalue gets the value 0x1A000000!!!!!
> How is this possible? Should'nt it be 0x34000000???
>
> Best Regards
> Jannick
>
>

hongky, Oct 9, 2003
5. ### Dan PopGuest

In <bm3cko\$qan\$-c.dk> "Jannick" <> writes:

>unsigned int myvalue=0;
>unsigned char mytest=0;
>
>mytest=0x34;
>myvalue = mytest<<24;
>
>Then myvalue gets the value 0x1A000000!!!!!
>How is this possible? Should'nt it be 0x34000000???

Show us a minimal, but *complete* program illustrating your problem.
Without seeing your code, not even my crystal ball is of much help here.
Except for suggesting that the shift count was actually 23 in your real
program.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email:

Dan Pop, Oct 9, 2003
6. ### peteGuest

Jannick wrote:
>
> Hi I have:
> unsigned int myvalue=0;
> unsigned char mytest=0;
>
> mytest=0x34;
> myvalue = mytest<<24;
>
> Then myvalue gets the value 0x1A000000!!!!!
> How is this possible? Should'nt it be 0x34000000???

I guess we're all assuming that the width of unsigned,
is at least 25 bits.
Is it, on your system ?

--
pete

pete, Oct 9, 2003
7. ### Robert StankowicGuest

"pete" <> schrieb im Newsbeitrag
news:...
> Jannick wrote:
> >
> > Hi I have:
> > unsigned int myvalue=0;
> > unsigned char mytest=0;
> >
> > mytest=0x34;
> > myvalue = mytest<<24;
> >
> > Then myvalue gets the value 0x1A000000!!!!!
> > How is this possible? Should'nt it be 0x34000000???

>
> I guess we're all assuming that the width of unsigned,
> is at least 25 bits.
> Is it, on your system ?

Well, if myvalue becomes 0x1A000000 we can assume that, can we?
And besides that, even if the width would be less than 24 the result would
be well defined and definitely not 0x1a000000
If I am not mistaken
on all sizes up to and including 26 bits the value would be 0x0,
27 and 28 bits will give 0x4000000,
29 bits 0x14000000,
30 bits and above 0x34000000

kind regards
Robert

Robert Stankowic, Oct 9, 2003
8. ### Dan PopGuest

In <3f856d52\$0\$32136\$> "Robert Stankowic" <> writes:

>"pete" <> schrieb im Newsbeitrag
>news:...
>> Jannick wrote:
>> >
>> > Hi I have:
>> > unsigned int myvalue=0;
>> > unsigned char mytest=0;
>> >
>> > mytest=0x34;
>> > myvalue = mytest<<24;
>> >
>> > Then myvalue gets the value 0x1A000000!!!!!
>> > How is this possible? Should'nt it be 0x34000000???

>>
>> I guess we're all assuming that the width of unsigned,
>> is at least 25 bits.
>> Is it, on your system ?

>
>Well, if myvalue becomes 0x1A000000 we can assume that, can we?

We can safely assume 29 bits. Which makes a minimum of 32 bits a
reasonable assumption. Pete should have engaged his brain before posting.

>And besides that, even if the width would be less than 24 the result would
>be well defined and definitely not 0x1a000000
>If I am not mistaken
>on all sizes up to and including 26 bits the value would be 0x0,
>27 and 28 bits will give 0x4000000,
>29 bits 0x14000000,
>30 bits and above 0x34000000

Wrong.

If the value of the right operand is negative or is greater than or
equal to the width in bits of the promoted left operand, the behavior
is undefined.

There was a long thread on this issue in comp.std.c, a few weeks ago.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email:

Dan Pop, Oct 9, 2003
9. ### peteGuest

Dan Pop wrote:
>
> In <3f856d52\$0\$32136\$> "Robert Stankowic" <> writes:
>
> >"pete" <> schrieb im Newsbeitrag
> >news:...
> >> Jannick wrote:
> >> >
> >> > Hi I have:
> >> > unsigned int myvalue=0;
> >> > unsigned char mytest=0;
> >> >
> >> > mytest=0x34;
> >> > myvalue = mytest<<24;
> >> >
> >> > Then myvalue gets the value 0x1A000000!!!!!
> >> > How is this possible? Should'nt it be 0x34000000???
> >>
> >> I guess we're all assuming that the width of unsigned,
> >> is at least 25 bits.
> >> Is it, on your system ?

> >
> >Well, if myvalue becomes 0x1A000000 we can assume that, can we?

>
> We can safely assume 29 bits. Which makes a minimum of 32 bits a
> reasonable assumption.
> Pete should have engaged his brain before posting.

A printf call could display 0x1A000000 with 16 bits and UB.

--
pete

pete, Oct 10, 2003
10. ### Robert StankowicGuest

"Dan Pop" <> schrieb im Newsbeitrag
news:bm3u7i\$12l\$...
> In <3f856d52\$0\$32136\$> "Robert

Stankowic" <> writes:
>
>
> >"pete" <> schrieb im Newsbeitrag
> >news:...
> >> Jannick wrote:
> >> >
> >> > Hi I have:
> >> > unsigned int myvalue=0;
> >> > unsigned char mytest=0;
> >> >
> >> > mytest=0x34;
> >> > myvalue = mytest<<24;
> >> >
> >> > Then myvalue gets the value 0x1A000000!!!!!
> >> > How is this possible? Should'nt it be 0x34000000???
> >>
> >> I guess we're all assuming that the width of unsigned,
> >> is at least 25 bits.
> >> Is it, on your system ?

> >
> >Well, if myvalue becomes 0x1A000000 we can assume that, can we?

>
> We can safely assume 29 bits. Which makes a minimum of 32 bits a
> reasonable assumption. Pete should have engaged his brain before posting.
>
> >And besides that, even if the width would be less than 24 the result

would
> >be well defined and definitely not 0x1a000000
> >If I am not mistaken
> >on all sizes up to and including 26 bits the value would be 0x0,
> >27 and 28 bits will give 0x4000000,
> >29 bits 0x14000000,
> >30 bits and above 0x34000000

>
> Wrong.
>
> If the value of the right operand is negative or is greater than or
> equal to the width in bits of the promoted left operand, the behavior
> is undefined.
>

So I obviously misunderstand the text from N869:

"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."

The semicolon and the lower case "otherwise" suggested to me that the last
phrase belongs to the sentence about signed types.
Thank you for the clarification.
kind regards
Robert

Robert Stankowic, Oct 10, 2003
11. ### JannickGuest

"Dan Pop" <> wrote in message
news:bm3k7e\$mu7\$...
> In <bm3cko\$qan\$-c.dk> "Jannick" <> writes:
>
> >unsigned int myvalue=0;
> >unsigned char mytest=0;
> >
> >mytest=0x34;
> >myvalue = mytest<<24;
> >
> >Then myvalue gets the value 0x1A000000!!!!!
> >How is this possible? Should'nt it be 0x34000000???

>
> Show us a minimal, but *complete* program illustrating your problem.
> Without seeing your code, not even my crystal ball is of much help here.
> Except for suggesting that the shift count was actually 23 in your real
> program.
>
> Dan
> --
> Dan Pop
> DESY Zeuthen, RZ group
> Email:

Here is my code. I would actually like to do the commented operation, but
also with the code as it is it gives me the strange error :-(. The funny
thing is that when I do measuredata.n1.number1<<16 I get ram_value =
0x00340000, but when I do measuredata.n1.number1<<24 I get 0x1A000000. It
seems to me that it actually only does measuredata.n1.number1<<23....I can
seem to see why :-(

Code:

struct no1_{
unsigned char number1;
unsigned char number2;
};

struct no2_{
unsigned int number3;
unsigned int number4;
};

struct measurement_data{
struct no1_ n1;
struct no2_ n2;
}measuredata;

int main(){

unsigned int* sram_location = (unsigned int *) 0xFF120011;

measuredata.n1.number1 = 0x34;
measuredata.n1.number2 = 0xBA;
measuredata.n2.number3 = 0x12345678;
measuredata.n2.number4 = 0xABCDEF12;
mytest= 0x34;

/* *sram_location = ( ((measuredata.n1.number1)<<24) | (
(measuredata.n1.number2)<<16) | ( (measuredata.n2.number3)>>16) );*/

ram_value = measuredata.n1.number1<<24;
*sram_location = ram_value;

return 0;
}

Jannick, Oct 10, 2003
12. ### Dan PopGuest

In <> pete <> writes:

>Dan Pop wrote:
>>
>> In <3f856d52\$0\$32136\$> "Robert Stankowic" <> writes:
>>
>> >"pete" <> schrieb im Newsbeitrag
>> >news:...
>> >> Jannick wrote:
>> >> >
>> >> > Hi I have:
>> >> > unsigned int myvalue=0;
>> >> > unsigned char mytest=0;
>> >> >
>> >> > mytest=0x34;
>> >> > myvalue = mytest<<24;
>> >> >
>> >> > Then myvalue gets the value 0x1A000000!!!!!
>> >> > How is this possible? Should'nt it be 0x34000000???
>> >>
>> >> I guess we're all assuming that the width of unsigned,
>> >> is at least 25 bits.
>> >> Is it, on your system ?
>> >
>> >Well, if myvalue becomes 0x1A000000 we can assume that, can we?

>>
>> We can safely assume 29 bits. Which makes a minimum of 32 bits a
>> reasonable assumption.
>> Pete should have engaged his brain before posting.

>
>A printf call could display 0x1A000000 with 16 bits and UB.

The probability of this happening is so close to that of nasal demon
generation that it's not worth considering in the context of a *real*
implementation.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email:

Dan Pop, Oct 10, 2003
13. ### Dan PopGuest

In <3f861fe8\$0\$32136\$> "Robert Stankowic" <> writes:

>"Dan Pop" <> schrieb im Newsbeitrag
>news:bm3u7i\$12l\$...
>> In <3f856d52\$0\$32136\$> "Robert

>Stankowic" <> writes:
>>
>>
>> >"pete" <> schrieb im Newsbeitrag
>> >news:...
>> >> Jannick wrote:
>> >> >
>> >> > Hi I have:
>> >> > unsigned int myvalue=0;
>> >> > unsigned char mytest=0;
>> >> >
>> >> > mytest=0x34;
>> >> > myvalue = mytest<<24;
>> >> >
>> >> > Then myvalue gets the value 0x1A000000!!!!!
>> >> > How is this possible? Should'nt it be 0x34000000???
>> >>
>> >> I guess we're all assuming that the width of unsigned,
>> >> is at least 25 bits.
>> >> Is it, on your system ?
>> >
>> >Well, if myvalue becomes 0x1A000000 we can assume that, can we?

>>
>> We can safely assume 29 bits. Which makes a minimum of 32 bits a
>> reasonable assumption. Pete should have engaged his brain before posting.
>>
>> >And besides that, even if the width would be less than 24 the result

>would
>> >be well defined and definitely not 0x1a000000
>> >If I am not mistaken
>> >on all sizes up to and including 26 bits the value would be 0x0,
>> >27 and 28 bits will give 0x4000000,
>> >29 bits 0x14000000,
>> >30 bits and above 0x34000000

>>
>> Wrong.
>>
>> If the value of the right operand is negative or is greater than or
>> equal to the width in bits of the promoted left operand, the behavior
>> is undefined.
>>

>
>So I obviously misunderstand the text from N869:
>
>"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."
>
>The semicolon and the lower case "otherwise" suggested to me that the last
>phrase belongs to the sentence about signed types.
>Thank you for the clarification.

3 *before* reading paragraph 4. Paragraph 3 contains the text I have
quoted (actually a similar text, since I was quoting from C89).

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email:

Dan Pop, Oct 10, 2003
14. ### Dan PopGuest

In <bm646n\$ail\$-c.dk> "Jannick" <> writes:

>Here is my code. I would actually like to do the commented operation, but
>also with the code as it is it gives me the strange error :-(. The funny
>thing is that when I do measuredata.n1.number1<<16 I get ram_value =
>0x00340000, but when I do measuredata.n1.number1<<24 I get 0x1A000000. It
>seems to me that it actually only does measuredata.n1.number1<<23....I can
>seem to see why :-(
>
>Code:
>
>struct no1_{
> unsigned char number1;
> unsigned char number2;
>};
>
>struct no2_{
> unsigned int number3;
> unsigned int number4;
>};
>
>struct measurement_data{
> struct no1_ n1;
> struct no2_ n2;
>}measuredata;
>
>
>int main(){
>
> unsigned int* sram_location = (unsigned int *) 0xFF120011;

^^^^^^^^^^
This is a *very* suspicious address for an unsigned int! Are you sure
you know what you're doing?

> measuredata.n1.number1 = 0x34;
> measuredata.n1.number2 = 0xBA;
> measuredata.n2.number3 = 0x12345678;
> measuredata.n2.number4 = 0xABCDEF12;
> mytest= 0x34;
>
>
> /* *sram_location = ( ((measuredata.n1.number1)<<24) | (
>(measuredata.n1.number2)<<16) | ( (measuredata.n2.number3)>>16) );*/
>
> ram_value = measuredata.n1.number1<<24;
> *sram_location = ram_value;
>
>return 0;
>}

Nope, this is NOT your code! It contains two undeclared identifiers,
mytest and ram_value and it generates no output, so that we can actually
see what happens. If I remove mytest, declare ram_value as unsigned int
and display its value instead of dereferencing a dubious pointer, I get
the following result:

fangorn:~/tmp 544> gcc test.c
fangorn:~/tmp 545> ./a.out
34000000

So, please don't waste our time with bogus pieces of code that neither

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email:

Dan Pop, Oct 10, 2003
15. ### CBFalconerGuest

Jannick wrote:
> "Dan Pop" <> wrote in message
>

.... snip ...
> >
> > Show us a minimal, but *complete* program illustrating your problem.
> > Without seeing your code, not even my crystal ball is of much help here.
> > Except for suggesting that the shift count was actually 23 in your real
> > program.

>
> Here is my code. I would actually like to do the commented operation, but
> also with the code as it is it gives me the strange error :-(. The funny
> thing is that when I do measuredata.n1.number1<<16 I get ram_value =
> 0x00340000, but when I do measuredata.n1.number1<<24 I get 0x1A000000. It
> seems to me that it actually only does measuredata.n1.number1<<23....I can
> seem to see why :-(
>
> Code:
>
> struct no1_{
> unsigned char number1;
> unsigned char number2;
> };
>
> struct no2_{
> unsigned int number3;
> unsigned int number4;
> };
>
> struct measurement_data{
> struct no1_ n1;
> struct no2_ n2;
> }measuredata;
>
> int main(){
>
> unsigned int* sram_location = (unsigned int *) 0xFF120011;

Undefined behavior here.

>
> measuredata.n1.number1 = 0x34;
> measuredata.n1.number2 = 0xBA;
> measuredata.n2.number3 = 0x12345678;
> measuredata.n2.number4 = 0xABCDEF12;
> mytest= 0x34;
>
> /* *sram_location = ( ((measuredata.n1.number1)<<24) | (
> (measuredata.n1.number2)<<16) | ( (measuredata.n2.number3)>>16) );*/
>
> ram_value = measuredata.n1.number1<<24;
> *sram_location = ram_value;
>
> return 0;
> }

Not a complete program. No way of displaying any fault. Illegal
constructs. Why are you wasting our time with this?

--
Chuck F () ()
Available for consulting/temporary embedded and systems.

CBFalconer, Oct 10, 2003
16. ### Default UserGuest

CBFalconer wrote:

> > unsigned int* sram_location = (unsigned int *) 0xFF120011;

>
> Undefined behavior here.

Mmmmm, nope.

[#5] An integer may be converted to any pointer type. |
Except as previously specified, the result is
implementation-defined, might not be properly aligned, and
might not point to an entity of the referenced type.49)

49)The mapping functions for converting a pointer to an
integer or an integer to a pointer are intended to be
consistent with the addressing structure of the execution
environment.

Assignment is ok. Use of it is problematic.

Brian Rodenborn

Default User, Oct 10, 2003
17. ### JannickGuest

"Default User" <> wrote in message
news:...
> CBFalconer wrote:
>
> > > unsigned int* sram_location = (unsigned int *) 0xFF120011;

> >
> > Undefined behavior here.

>
>
> Mmmmm, nope.
>
>
> [#5] An integer may be converted to any pointer type. |
> Except as previously specified, the result is
> implementation-defined, might not be properly aligned, and
> might not point to an entity of the referenced type.49)
>
>
> 49)The mapping functions for converting a pointer to an
> integer or an integer to a pointer are intended to be
> consistent with the addressing structure of the execution
> environment.
>
>
> Assignment is ok. Use of it is problematic.
>
>
>
> Brian Rodenborn

Sorry to post defective code....I was to quick in cutting the problem out of
entire code... I think it is my compiler tricking me :-(, the "unsigned int*
sram_location = (unsigned int *) 0xFF120011" is okay because the code is
intended for a microcontroller that has a peripheral device memory-mapped at
this location.

I solved the problem though, but do still not understand it...I think the
error lies in the compiler and dont think the error is easily seen. When I
did:

*sram_location = ( ((measuredata.n1.number1)<<24) | (
(measuredata.n1.number2)<<16) | ( (measuredata.n2.number3)>>16) );

I got the 0x1A000000

When I do the:
*sram_location = (
((measuredata.n1.number1<<16)<<8) | ((measuredata.n1.number2)<<16) | (
(measuredata.n2.number3) >>16) );
I get the correct result: 0x34000000

Thank you all for your time, effort and patience.

Best Regards
Jannick

Jannick, Oct 11, 2003
18. ### Robert StankowicGuest

"Dan Pop" <> schrieb im Newsbeitrag
news:bm67l8\$ekm\$...
> In <3f861fe8\$0\$32136\$> "Robert

Stankowic" <> writes:
>
>
> >"Dan Pop" <> schrieb im Newsbeitrag

[....]

> The origin of your misunderstanding is the omission of reading paragraph
> 3 *before* reading paragraph 4. Paragraph 3 contains the text I have
> quoted (actually a similar text, since I was quoting from C89).

Precise as always
Thank you
Robert

Robert Stankowic, Oct 11, 2003
19. ### CBFalconerGuest

Jannick wrote:
>

.... snip ...
>
> Sorry to post defective code....I was to quick in cutting the
> problem out of entire code... I think it is my compiler tricking
> me :-(, the "unsigned int* sram_location = (unsigned int *)
> 0xFF120011" is okay because the code is intended for a
> microcontroller that has a peripheral device memory-mapped at
> this location.
>
> I solved the problem though, but do still not understand it...I
> think the error lies in the compiler and dont think the error is
> easily seen. When I did:

If you are writing into a section of memory mapped i/o, some of
those locations are inputs. They can't be expected to store
values.

--
Chuck F () ()
Available for consulting/temporary embedded and systems.

CBFalconer, Oct 11, 2003
20. ### Keith ThompsonGuest

"Jannick" <> writes:
[...]
> Sorry to post defective code....I was to quick in cutting the problem out of
> entire code... I think it is my compiler tricking me :-(, the "unsigned int*
> sram_location = (unsigned int *) 0xFF120011" is okay because the code is
> intended for a microcontroller that has a peripheral device memory-mapped at
> this location.
>
> I solved the problem though, but do still not understand it...I think the
> error lies in the compiler and dont think the error is easily seen. When I
> did:
>
> *sram_location = ( ((measuredata.n1.number1)<<24) | (
> (measuredata.n1.number2)<<16) | ( (measuredata.n2.number3)>>16) );
>
> I got the 0x1A000000
>
> When I do the:
> *sram_location = (
> ((measuredata.n1.number1<<16)<<8) | ((measuredata.n1.number2)<<16) | (
> (measuredata.n2.number3) >>16) );
> I get the correct result: 0x34000000
>

The only difference (which would have been a lot clearer if you had
formatted and indented the code more clearly) is the change from

(measuredata.n1.number1)<<24

to

(measuredata.n1.number1<<16)<<8

What happens when you display the values of those expressions in
isolation? What are the types of number1, number2, number3, and
*sram_location? What is the size of int on your platform?

If number1 is an unsigned int, and unsigned int is 16 bits on your
platform, you could be seeing undefined behavior because the right
operand of the "<<" is greater than or equal to the size of the
operand. If this is the problem, and you're trying to get an unsigned
long result (at least 32 bits), you should cast each operand to unsigned
long before shifting it:

*sram_location =
( (unsigned long)measuredata.n1.number1 << 24 ) |
( (unsigned long)measuredata.n1.number2 << 16 ) |
( (unsigned long)measuredata.n2.number3 >> 16 );

(On the other hand, if number3 is 16 bits, it doesn't make much sense
to do a 16-bit right shift on it.)

--
Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
Schroedinger does Shakespeare: "To be *and* not to be"

Keith Thompson, Oct 11, 2003