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da
#1 / 22
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 casting of struct element void pointer casting Hi
I am having trouble with a following line of C code:
((char *)structure_name.void_pointer)-=3;
Error: lvalue required
Thanks for any help,
Dan
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| Tue, 27 Sep 2005 04:45:49 GMT |
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Mike Wahle
#2 / 22
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casting of struct element void pointer casting
Quote: > Hi > I am having trouble with a following line of C code:
> ((char *)structure_name.void_pointer)-=3;
at least 3 bytes past the address of the object
or array to which it intitially pointed? And
does that object or array have a size of at
least 3? If not, I believe that's UB.
That cast should also imo probably be to 'unsigned char*'
Quote: (cast-to-type)(object)
.. creates a temporary object of the casted-to type.
Temporaries are not lvalues. Try something like:
unsigned char *p = structure_name.void_pointer;
p -= 3;
structure_name.void_pointer = p;
Also you should probably try to change that literal
3 to a macro or the result of calculating an object
or array size.
What specifically is the statement you posted
supposed to do?
-Mike
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| Tue, 27 Sep 2005 05:10:32 GMT |
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Thomas Stege
#3 / 22
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casting of struct element void pointer casting Quote:
> Hi > I am having trouble with a following line of C code:
> ((char *)structure_name.void_pointer)-=3;
> Error: lvalue required
An lvalue means that an expression which denotes an object
in memory is needed. If you have a variable called var then
in
var = 1; /*for suitable types of var*/
var denotes an object in memory.
You cannot do this however
1 = 3;
For obvious reasons.
The cast operator returns a values, not something that denotes an
object so what you are doing above is somewhat analogous to 1 = 3;
(it is probably closer to (var) = 3;, but the idea is the same).
If you are really wanting to do pointer arithmetic then something
more fundamental is wrong with your design I would say. If you are
trying to subtract 3 from a char then you are missing a * operator
in there.
*((char *)structure_name.void_pointer)-=3;
Might or might not be what you want. You aren't exactly giving much
information so it is hard to tell.
HTH.
--
Thomas.
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| Tue, 27 Sep 2005 05:39:30 GMT |
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Thomas Stege
#4 / 22
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casting of struct element void pointer casting Quote:
>>Error: lvalue required > (cast-to-type)(object)
> .. creates a temporary object of the casted-to type.
It just returns a value surely. No objects are created.
--
Thomas.
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| Tue, 27 Sep 2005 05:44:11 GMT |
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Ben Pfaf
#5 / 22
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casting of struct element void pointer casting Quote:
> ((char *)structure_name.void_pointer)-=3; > Error: lvalue required
The cast operator does not yield an lvalue. You can only assign
to an lvalue. If structure_name.void_pointer has type `void *',
then you can do something equivalent with
structure_name.void_pointer
= ((char *) structure_name.void_pointer) - 3;
I can't recall right now whether the parentheses around the cast
expression are necessary, but they won't hurt anything.
--
"To get the best out of this book, I strongly recommend that you read it."
--Richard Heathfield
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| Tue, 27 Sep 2005 06:48:53 GMT |
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Mike Wahle
#6 / 22
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casting of struct element void pointer casting
Quote:
> >>Error: lvalue required
> > (cast-to-type)(object)
> > .. creates a temporary object of the casted-to type.
> It just returns a value surely. No objects are created.
I'm using the term 'temporary object' (perhaps
incorrectly) where you're using 'value'.
-Mike
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| Tue, 27 Sep 2005 08:00:44 GMT |
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Kevin Easto
#7 / 22
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casting of struct element void pointer casting Quote:
>> ((char *)structure_name.void_pointer)-=3;
>> Error: lvalue required
> The cast operator does not yield an lvalue. You can only assign
> to an lvalue. If structure_name.void_pointer has type `void *',
> then you can do something equivalent with
> structure_name.void_pointer
> = ((char *) structure_name.void_pointer) - 3;
> I can't recall right now whether the parentheses around the cast
> expression are necessary, but they won't hurt anything.
They're not - all the unary operators (cast included) have higher
precedence than the mathematical operators (a fairly easy rule of thumb
to remember).
- Kevin.
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| Tue, 27 Sep 2005 10:16:58 GMT |
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Alex Russel
#8 / 22
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casting of struct element void pointer casting
Quote: > Hi > I am having trouble with a following line of C code:
> ((char *)structure_name.void_pointer)-=3;
> Error: lvalue required
> Thanks for any help,
> Dan
I assume it is something like:
struct test
{
void *void_pointer;
Quote: };
struct test structure_name;
( (char *) structure_name.void_point)-=3;
--
Alex Russell
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| Tue, 27 Sep 2005 14:04:31 GMT |
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Ben Pfaf
#9 / 22
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casting of struct element void pointer casting Quote:
> it would help a lot to see the declaration for structure_name.void_pointer > I assume it is something like: > struct test
> {
> void *void_pointer;
> };
> struct test structure_name;
> ( (char *) structure_name.void_point)-=3;
I hope you're not suggesting that that could be valid code. It's
not: the cast operator does not yield an lvalue, so its result
cannot be used as the left side of an assignment operator.
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}
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| Tue, 27 Sep 2005 14:06:41 GMT |
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dac
#10 / 22
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casting of struct element void pointer casting
Quote:
>> ((char *)structure_name.void_pointer)-=3;
>> Error: lvalue required
> The cast operator does not yield an lvalue. You can only assign
> to an lvalue. If structure_name.void_pointer has type `void *',
> then you can do something equivalent with
> structure_name.void_pointer
> = ((char *) structure_name.void_pointer) - 3;
> I can't recall right now whether the parentheses around the cast
> expression are necessary, but they won't hurt anything.
Thank you, I did not know that casting could not make lvalue
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| Tue, 27 Sep 2005 16:39:10 GMT |
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dac
#11 / 22
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casting of struct element void pointer casting
Quote: > *((char *)structure_name.void_pointer)-=3;
> Might or might not be what you want. You aren't exactly giving much
> information so it is hard to tell.
> HTH.
I am trying to get void * to point to 3 character sizes below its current
pointing postion, not take 3 from the character at the current pointer
position.
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| Tue, 27 Sep 2005 16:43:08 GMT |
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CBFalcone
#12 / 22
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casting of struct element void pointer casting Quote:
> > *((char *)structure_name.void_pointer)-=3;
> > Might or might not be what you want. You aren't exactly giving
> > much information so it is hard to tell.
> I am trying to get void * to point to 3 character sizes below its
> current pointing postion, not take 3 from the character at the
> current pointer position.
char *p;
...
p = structure_name.void_pointer; /* no cast */
p -= 3;
assuming the void_pointer points somewhere within a suitable
entity. If it is the result of a malloc, for example, this is not
necessarily valid, but it is the kind of thing done in system code
in accessing system fields during malloc/new/realloc operations.
You can see an example (for djgpp at least) in:
<http://cbfalconer.home.att.net/download/nmalloc.zip>
--
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net> USE worldnet address!
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| Tue, 27 Sep 2005 18:32:37 GMT |
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da
#13 / 22
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casting of struct element void pointer casting Quote: > The cast operator returns a values, not something that denotes an > object so what you are doing above is somewhat analogous to 1 = 3; > (it is probably closer to (var) = 3;, but the idea is the same). > If you are really wanting to do pointer arithmetic then something > more fundamental is wrong with your design I would say. If you are > trying to subtract 3 from a char then you are missing a * operator > in there. > *((char *)structure_name.void_pointer)-=3;
> Might or might not be what you want. You aren't exactly giving much
> information so it is hard to tell.
> HTH.
I am trying to get the pointer to point to 3 character spaces below
its current value.
P.S.
Is there a previous C standard where the originally posted line would
have worked? I am updating C code to a new compiler and this is one
line that would not compile.
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| Tue, 27 Sep 2005 18:57:34 GMT |
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da
#14 / 22
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casting of struct element void pointer casting Quote: > The cast operator returns a values, not something that denotes an > object so what you are doing above is somewhat analogous to 1 = 3; > (it is probably closer to (var) = 3;, but the idea is the same). > If you are really wanting to do pointer arithmetic then something > more fundamental is wrong with your design I would say. If you are > trying to subtract 3 from a char then you are missing a * operator > in there. > *((char *)structure_name.void_pointer)-=3;
> Might or might not be what you want. You aren't exactly giving much
> information so it is hard to tell.
> HTH.
I am trying to get the pointer to point to 3 character spaces below
its current address.
P.S.
Is there a previous C standard where the originally posted line would
have worked? I am updating C code to a new compiler and this is one
line that would not compile.
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| Tue, 27 Sep 2005 18:57:36 GMT |
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da
#15 / 22
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casting of struct element void pointer casting Quote: > The cast operator does not yield an lvalue. You can only assign
> to an lvalue. If structure_name.void_pointer has type `void *',
> then you can do something equivalent with
> structure_name.void_pointer
> = ((char *) structure_name.void_pointer) - 3;
> I can't recall right now whether the parentheses around the cast
> expression are necessary, but they won't hurt anything.
Thank you,
I did not know that cast operator not make lvalue.
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| Tue, 27 Sep 2005 18:58:08 GMT |
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