Is the following code conformat to ANSI C?
typedef struct {
int a;
int b;
} doomdata;
int main(void)
{
int x;
x = (int)&((doomdat a*)0)->b;
printf("x=%d\n" , x);
return x;
}
The part,
(int)&((doomdat a*)0)->b;
is it conformant to ANSI C? What is it supposed to do?
Thanks for any tips.
Napi
-- http://www.axiomsol.com http://www.cs.indiana.edu/hyplan/napi.html
Nov 14 '05
60  2965 
"Malcolm" <ma*****@55bank .freeserve.co.u k> wrote in message
news:co******** **@newsg3.svr.p ol.co.uk... "Christian Bau" <ch***********@ cbau.freeserve. co.uk> wrote
In C99, there seems to be an actual requirement that casting an integer
zero to a pointer type will produce a null pointer. If null pointers
don't have all bits zero, and an integer 0 has all bits zero, and
converting an integer 0 to a pointer produces a null pointer, then logic
says that this conversion cannot leave the bits unchanged.
But in this case we are not casting an integer 0 to a pointer, but a null
pointer to an integer.
No, you're not.
You casting integer 0 into a null pointer, offsetting
that pointer by the distance to the member field, then
taking the address of that field (which converts the
pointer back to an integer -- that reverses whatever
happened in converting the integer to a pointer). That
leaves a simple integer constant that can be resolved
at compile time.
The & cancels out the ->. There is no load of a pointer
into an address register whatsoever and no runtime
dereference.
"Malcolm" <ma*****@55bank .freeserve.co.u k> wrote: "Jack Klein" <ja*******@spam cop.net> wrote
When will people get it through their heads that it makes not a whit
of difference whether all bits zero happens to be a representation of
a null pointer on a particular platform?
Also consider if a null pointer is cast to an integral type, or if a pointer
derived by adding an offset to the null pointer is cast to an integer. This
cast is a simple bitwise conversion, so results will differ on a platform on
which NULL is not all bits zero.
Actually the cast is implementation-defined (and the implementation
can define it to be UB in some, or all, cases).
It would make sense for an implementation to give 0 for
(int)(void *)0, even if (void *)0 were not all bits zero.
The expression (int)&((doomdat a*)0)->b is relying on 2 non-standard
things:
- the compiler doesn't actually dereference 0
- the cast to int works as if pointers are ints in a flat
memory model and (int)NULL == 0
It's irrelevant to this expression whether NULL is all-bits-zero
or not.
"Jack Klein" <ja*******@spam cop.net> wrote in message
news:bn******** *************** *********@4ax.c om... On 28 Nov 2004 20:29:37 GMT, "S.Tobias"
<si***@FamOuS.B edBuG.pAlS.INVA LID> wrote in comp.lang.c:
Jack Klein <ja*******@spam cop.net> wrote: On 28 Nov 2004 10:41:21 GMT, "S.Tobias"
<si***@FamOuS.B edBuG.pAlS.INVA LID> wrote in comp.lang.c:
> Mohd Hanafiah Abdullah <na**@cs.indian a.edu> wrote: > > typedef struct {
> > int a;
> > int b;
> > } doomdata;
> Would this be correct?
> (int)&((doomdat a*)0)->a;
Technically it is still undefined behavior, as the semantics of the
expression dereference a null pointer.
(This deserves a separate thread, but since I asked the above
question here, I'll continue here too.)
As I understand the expression constitutes an access to the structure
member.
1. Does a member access constitute an access to the *whole* structure?
eg.:
The term 'access' is really only used in the C standard in conjunction
with the volatile qualifier, where the wording is unfortunately vague
enough that it can be construed several different ways.
struct A { int i; int _i; };
struct B { int i; float f; };
struct A a = {0};
struct B *pb = (struct B*)&a;
pb->i; //UB?
(*pb).i; //UB?
The two statements above actually have defined behavior, but not for
the reason you might think. The language guarantees that a pointer to
structure, suitably cast, is also a pointer to its first member. So
in given that 'pb' holds any of the following:
- a pointer to any structure type whose first member is an int
- a pointer to an array of ints
- a pointer to a single int
...and the int pointed to has a valid value, the expressions, though
not recommended, will work as designed. The compiler must generate
code equivalent to *(int *)pb, and if there is actually an int there
all is well.
Do I access the first int sub-object in `a' only, or do I access
the whole object `a'?
Now that depends what you mean by access. Let's assume a Pentium
(1/2/3/4) platform with a typical compiler, which means the size of
either of your structures is 8 8-bit bytes. Now let's also assume
that this implementation allocates all structures on an address evenly
divisible by 8, a not uncommon performance feature of such
implementations .
With the assumptions above, if the processor needs to access the
structure from memory it will perform a 64-bit access physically, so
even though your code does not direct the abstract machine to touch
the value of other members in any way, the entire memory space holding
the structure will be physically read.
2. I see certain similarity between structs and arrays (in fact,
both are called "aggregates ").
Why is it that for array:
&a[5];
doesn't constitute object access (6.5.3.2#3), whereas for struct:
&s.m;
&ps->m;
the expressions do constitute access?
Why is the language designed like this?
There are actually more differences than similarities between structs
and arrays, despite the fact that both are aggregates. Structs are
first class objects, meaning they can be assigned, passed to and
returned from functions by value, and their names are never implicitly
converted to pointers. Arrays are not first class objects and do not
share any of the characteristics above.
As for other differences in this particular case, this is spelled out
by paragraph 3 of 6.5.3.2 of C99:
[begin quotation]
The unary & operator returns the address of its operand. If the
operand has type ''type'', the result has type ''pointer to type''. If
the operand is the result of a unary * operator, neither that operator
nor the & operator is evaluated and the result is as if both were
omitted, except that the constraints on the operators still apply and
the result is not an lvalue. Similarly, if the operand is the result
of a [] operator, neither the & operator nor the unary * that is
implied by the [] is evaluated and the result is as if the & operator
were removed and the [] operator were changed to a + operator.
Otherwise, the result is a pointer to the object or function
designated by its operand.
[end quotation]
Note the differences between applying '&' to the result of a '*'
operator and to the result of a '[]' operator. In the former case,
neither '&' nor '*' are evaluated as such, but note "the constraints
on the operators still apply".
Now let's back up to paragraph 1 of 6.5.3.2, which lists the
constraints for the unary '&' operator:
[begin quotation]
The operand of the unary & operator shall be either a function
designator, the result of a [] or unary * operator, or an lvalue that
designates an object that is not a bit-field and is not declared with
the register storage-class specifier.
[end quotation]
Notice that the expression under discussion,
(int)&((doomdat a*)0)->a;
...is none of these things. Specifically, the operand of the '&'
operator, '((doomdata*)0)->a' is:
- not a function designator
- not the result of a [] operator
- not the result of a unary * operator
Yes it is unary *. a->b is an alias for (*a).b
- and, because of the null pointer, not an lvalue
Finally consider one last thing, namely that regardless of whether
there is an actual access to an object, the expression explicitly
performs pointer arithmetic on a null pointer, and such use of a null
pointer is undefined in and of itself.
Therefore, & cancels out the -> to yield a simple
integer constant that is resolved at compile time.
In article <q1************ *****@newsread3 .news.pas.earth link.net>
xarax <xa***@email.co m> wrote: ... the compiler will not generate a load into an address register,
because the & cancels out the ->. The compiler has all the information
it needs to resolve the expression to a constant offset value.
There is only one problem: the compiler really sucks. It does
only what it is absolutely forced to by the C Standard. The C
Standard *allows* it to follow the pointer first, and only then
compute the offset, so it does.
(Well, which "the" compiler are *you* talking about?)
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.
"Chris Torek" <no****@torek.n et> wrote in message
news:co******** *@news2.newsguy .com... In article <q1************ *****@newsread3 .news.pas.earth link.net>
xarax <xa***@email.co m> wrote:... the compiler will not generate a load into an address register,
because the & cancels out the ->. The compiler has all the information
it needs to resolve the expression to a constant offset value.
There is only one problem: the compiler really sucks. It does
only what it is absolutely forced to by the C Standard. The C
Standard *allows* it to follow the pointer first, and only then
compute the offset, so it does.
Following the pointer into memory would make it
impossible to determine the address of the field.
The & cancels out the apparent dereference.
>> In article <q1************ *****@newsread3 .news.pas.earth link.net> xarax <xa***@email.co m> wrote: >... the compiler will not generate a load into an address register,
>because the & cancels out the ->. The compiler has all the information
>it needs to resolve the expression to a constant offset value.
"Chris Torek" <no****@torek.n et> wrote in message
news:co******* **@news2.newsgu y.com... There is only one problem: the compiler really sucks. It does
only what it is absolutely forced to by the C Standard. The C
Standard *allows* it to follow the pointer first, and only then
compute the offset, so it does.
In article <PU************ *****@newsread3 .news.pas.earth link.net>
xarax <xa***@email.co m> wrote:Following the pointer into memory would make it
impossible to determine the address of the field.
Yes. This is why the compiler throws the result away after following
the pointer.
The & cancels out the apparent dereference.
No, the "&" makes the compiler throw away the result of the dereference.
Unfortunately, by then it is too late.
I did say this compiler really sucks. But it conforms.
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.
"Chris Torek" <no****@torek.n et> wrote in message
news:co******** *@news2.newsguy .com... In article <q1************ *****@newsread3 .news.pas.earth link.net>
xarax <xa***@email.co m> wrote:
>... the compiler will not generate a load into an address register,
>because the & cancels out the ->. The compiler has all the information
>it needs to resolve the expression to a constant offset value.
"Chris Torek" <no****@torek.n et> wrote in message
news:co******* **@news2.newsgu y.com... There is only one problem: the compiler really sucks. It does
only what it is absolutely forced to by the C Standard. The C
Standard *allows* it to follow the pointer first, and only then
compute the offset, so it does.
In article <PU************ *****@newsread3 .news.pas.earth link.net>
xarax <xa***@email.co m> wrote:Following the pointer into memory would make it
impossible to determine the address of the field.
Yes. This is why the compiler throws the result away after following
the pointer.
The & cancels out the apparent dereference.
No, the "&" makes the compiler throw away the result of the dereference.
Unfortunately, by then it is too late.
I did say this compiler really sucks. But it conforms.
Totally ridiculous.
"xarax" <xa***@email.co m> wrote: "Malcolm" <ma*****@55bank .freeserve.co.u k> wrote: "Christian Bau" <ch***********@ cbau.freeserve. co.uk> wrote:
In C99, there seems to be an actual requirement that casting an integer
zero to a pointer type will produce a null pointer. If null pointers
don't have all bits zero, and an integer 0 has all bits zero, and
converting an integer 0 to a pointer produces a null pointer, then logic
says that this conversion cannot leave the bits unchanged.
But in this case we are not casting an integer 0 to a pointer, but a null
pointer to an integer.
No, you're not.
You casting integer 0 into a null pointer,
Casting 0 to a pointer must give a null pointer.
offsetting that pointer by the distance to the member
field, then taking the address of that field
(which converts the pointer back to an integer
That conversion is implementation-defined
-- that reverses whatever happened in converting the
integer to a pointer).
There is no requirement for that to be true. In fact
it can't be true, if there is not an exact mapping
from integers to pointers (eg. segmented architecture,
IA64, etc.)
If you're still not convinced, imagine that NULL
lives at address 0xDEADBEEF. Then &((foo *)0)->bar
might be 0xDEADC00C, for example, and when that's
converted back to an integer you might get 0xDEADC00C
still. Not a very accurate struct offset.
Furthermore, if that is outside the range of signed int
(likely on a 32-bit system) you get undefined behaviour
(again).
That leaves a simple integer constant that can be resolved
at compile time.
There is no load of a pointer into an address register
whatsoever and no runtime dereference.
For your compiler on Tuesdays, perhaps. There is nothing
(except possible sales figures..) to stop a compiler from
loading 0 to an address register and then incrementing
it, causing a hardware exception.
In article <f3************ *************** *****@4ax.com>,
Jack Klein <ja*******@spam cop.net> wrote: On Sun, 28 Nov 2004 22:17:21 +0000, Christian Bau
<ch***********@ cbau.freeserve. co.uk> wrote in comp.lang.c:
In article <co**********@n ews7.svr.pol.co .uk>,
"Malcolm" <ma*****@55bank .freeserve.co.u k> wrote:
Also consider if a null pointer is cast to an integral type, or if a
pointer
derived by adding an offset to the null pointer is cast to an integer.
This
cast is a simple bitwise conversion, so results will differ on a platform
on
which NULL is not all bits zero.
Not quite; a conversion from a pointer type to an integer does whatever
the implementation thinks is a good idea; usually the bits of the
representation are copied unchanged, but that is not necessarily so.
In C99, there seems to be an actual requirement that casting an integer
zero to a pointer type will produce a null pointer. If null pointers
don't have all bits zero, and an integer 0 has all bits zero, and
converting an integer 0 to a pointer produces a null pointer, then logic
says that this conversion cannot leave the bits unchanged.
I am unsure of your meaning here. Do you mean an integral constant at
the source level, as in:
double *dp = (double * 0);
(double *) 0;
...or do you actually mean the value of an integer object, as in:
int x = 0;
double *dp = (double *x);
Both.
If you mean the latter, I have never noticed anything in C99 requiring
the result be a null pointer.
Could you please clarify and include chapter & verse?
Not directly. The result of conversion from int to for example double*
is defined directly in one special case: When the int is a null pointer
constant, that is the value is 0, and it is an integer constant
expression. But since the result of a conversion only depends on the
value converted and nothing else, the result of converting _any_ int of
value 0 must always be the same. As it is a null pointer in some cases,
it must be a null pointer in all cases.
Christian Bau <ch***********@ cbau.freeserve. co.uk> wrote: The result of conversion from int to for example double*
is defined directly in one special case: When the int is a null pointer
constant, that is the value is 0, and it is an integer constant
expression.
No, the value is not important here, you're misquoting the Standard.
The Standard says: "integer constant expression with the value 0",
so first you look up what is an integer constant expression, and
then you filter only those that have the value zero. Examples are:
0
0x0
0u
(1-1)
'\000'
(int)0.0
7/11
sizeof 13 - sizeof (int)
But since the result of a conversion only depends on the
value converted and nothing else, the result of converting _any_ int of
value 0 must always be the same. As it is a null pointer in some cases,
it must be a null pointer in all cases.
The Std does not say such a thing. If it wanted the value 0 to be
converted into a null pointer, it would say "integer expression with
the value 0" or just "integer value 0". Integer constant expression
is a special case that a compiler must recognize at compile-time.
Conversions between pointer and integer types are explicitly
implementation defined.
--
Stan Tobias
mailx `echo si***@FamOuS.Be dBuG.pAlS.INVALID | sed s/[[:upper:]]//g` This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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