«

I have couple of questions related to array addresses. As they belong
to the same block, I
am putting them here in one single post. I hope nobody minds:

char array[35];

int address;

Questions 1:
Why cannot I do the following:

address = (int)array;

whereas I found it perfectly alright to do the following:

address = (int)&array;

Question 2:
What is the difference between array and &array. Are not they the same
thing i.e. starting
address of the array.

Question 3:

Why cannot I do array++. Is array a pointer constant?

Question 4:
How can I do the following:

array = address;

i.e. Give a new starting address to the array. Is it possible?

Thanks in advance for all Your replies.

A.

"anonymous" <> wrote in message
news: ups.com...
>I have couple of questions related to array addresses. As they belong
> to the same block, I
> am putting them here in one single post. I hope nobody minds:
>
> char array[35];
>
> int address;
>
> Questions 1:
> Why cannot I do the following:
>
> address = (int)array;
>
> whereas I found it perfectly alright to do the following:
>
> address = (int)&array;
>
> Question 2:
> What is the difference between array and &array. Are not they the same
> thing i.e. starting
> address of the array.
>
> Question 3:
>
> Why cannot I do array++. Is array a pointer constant?
>
> Question 4:
> How can I do the following:
>
> array = address;
>
> i.e. Give a new starting address to the array. Is it possible?
>
> Thanks in advance for all Your replies.

>Questions 1:
>Why cannot I do the following:

>address = (int)array;

What do you by 'can't do'?


> Question 2:
> What is the difference between array and &array. Are not they the same
> thing i.e. starting

an array name decomposes into the address of its first element, i.e., a
constant. You obviously cannot take the address of a 'value', and so &array
doesn't make sense, and I assume the compiler's saying to itself 'well, ok,
I know what he means'.


> Question 3:
>
> Why cannot I do array++. Is array a pointer constant?

Because an array name decomposes into the address of its first element, so
it's like saying 42++;


> Question 4:
> How can I do the following:
>
> array = address;


You can't, it'd be like saying 42 = 24;

anonymous wrote:
> I have couple of questions related to array addresses. As they belong
> to the same block, I
> am putting them here in one single post. I hope nobody minds:
>
> char array[35];
>
> int address;
>
> Questions 1:
> Why cannot I do the following:
>
> address = (int)array;

No, it is a legal syntax. However pointer to integer conversion may
invoke undefined behavior if result cannot be represented in integer
type.

>
> whereas I found it perfectly alright to do the following:
>
> address = (int)&array;
>
> Question 2:
> What is the difference between array and &array. Are not they the same
> thing i.e. starting
> address of the array.

array and &array are same in value context, both yield pointer to
first element in value context.

>
> Question 3:
>
> Why cannot I do array++. Is array a pointer constant?

postfix increment operator needs a modifiable l-value. array name is
not a modifiable l-value.

>
> Question 4:
> How can I do the following:
>
> array = address;
>
> i.e. Give a new starting address to the array. Is it possible?

No, see above.

Krishanu

pemo wrote:
> "anonymous" <> wrote in message
> news: ups.com...
> >I have couple of questions related to array addresses. As they belong
> > to the same block, I
> > am putting them here in one single post. I hope nobody minds:
> >
> > char array[35];
> >
> > int address;
> >
> > Questions 1:
> > Why cannot I do the following:
> >
> > address = (int)array;
> >

> > whereas I found it perfectly alright to do the following:
> >
> > address = (int)&array;
> >
> > Question 2:
> > What is the difference between array and &array. Are not they the same
> > thing i.e. starting
> > address of the array.
> >
> > Question 3:
> >
> > Why cannot I do array++. Is array a pointer constant?
> >
> > Question 4:
> > How can I do the following:
> >
> > array = address;
> >
> > i.e. Give a new starting address to the array. Is it possible?
> >
> > Thanks in advance for all Your replies.
>
> >Questions 1:
> >Why cannot I do the following:
>
> >address = (int)array;
>
> What do you by 'can't do'?

Ooops, checking again, it is perfectly alright to do
address = (int)array;
My fault somewhere.
>
> > Question 2:
> > What is the difference between array and &array. Are not they the same
> > thing i.e. starting
>
> an array name decomposes into the address of its first element, i.e., a
> constant. You obviously cannot take the address of a 'value', and so &array
> doesn't make sense, and I assume the compiler's saying to itself 'well, ok,
> I know what he means'.

Why so liberty here when it keeps complaining all other times. I mean
the compiler
>
>
> > Question 3:
> >
> > Why cannot I do array++. Is array a pointer constant?
>
> Because an array name decomposes into the address of its first element, so
> it's like saying 42++;
>
>
> > Question 4:
> > How can I do the following:
> >
> > array = address;
>
>
> You can't, it'd be like saying 42 = 24;

Thanks for your help.

anonymous wrote:
> I have couple of questions related to array addresses. As they belong
> to the same block, I
> am putting them here in one single post. I hope nobody minds:
>
> char array[35];
>
> int address;
>
> Questions 1:
> Why cannot I do the following:
>
> address = (int)array;
>
> whereas I found it perfectly alright to do the following:
>
> address = (int)&array;
>
> Question 2:
> What is the difference between array and &array. Are not they the same
> thing i.e. starting
> address of the array.
>
> Question 3:
>
> Why cannot I do array++. Is array a pointer constant?
>
> Question 4:
> How can I do the following:
>
> array = address;
>
> i.e. Give a new starting address to the array. Is it possible?
>
> Thanks in advance for all Your replies.
>
> A.
>
Search this group for numerous discussion about hazels of converting int
to pointer.

Standard has added special type intptr_t and uintptr_t to enable integer
types to hold object pointers.
Discussed in section:
"7.18.1.4 Integer types capable of holding object pointers".

But this is a *nice to have*, your compiler can decide not implement
this data type.

Again, this is only for the conversion from/to void.


To answer your specific questions.
Q1: Both are undefined behavior

Q2: Both means the same i.e. address of the first element.
[ However the moment you pass an array to a function, the meaning
changes, and hence meaning of array and &array differs there ]

Q3&Q4: array is not an lvalue, see section:
6.3.2.1 Lvalues, arrays, and function designators
So you cannot modify it.
( Again the meaning of array changes the moment you pass it to
function)

--
(Welcome) http://www.ungerhu.com/jxh/clc.welcome.txt
(clc FAQ) http://www.eskimo.com/~scs/C-faq/top.html

pemo wrote:
> "anonymous" <> wrote in message
> news: ups.com...
[snip]
> > Question 2:
> > What is the difference between array and &array. Are not they the same
> > thing i.e. starting
>
> an array name decomposes into the address of its first element, i.e., a
> constant.
>
[snip]

Well, almost.

> You obviously cannot take the address of a 'value', and so &array
> doesn't make sense, and I assume the compiler's saying to itself 'well, ok,
> I know what he means'.

In a value context, an object of type "array N of T" becomes
a value of type "pointer to T", pointing to the first element
of that array, i.e., the one with subscript 0.

And do search the archives.

Anand wrote:
> anonymous wrote:
> > I have couple of questions related to array addresses. As they belong
> > to the same block, I
> > am putting them here in one single post. I hope nobody minds:
> >
> > char array[35];
> >
> > int address;
> >
> > Questions 1:
> > Why cannot I do the following:
> >
> > address = (int)array;
> >
> > whereas I found it perfectly alright to do the following:
> >
> > address = (int)&array;
> >
> > Question 2:
> > What is the difference between array and &array. Are not they the same
> > thing i.e. starting
> > address of the array.
> >
> > Question 3:
> >
> > Why cannot I do array++. Is array a pointer constant?
> >
> > Question 4:
> > How can I do the following:
> >
> > array = address;
> >
> > i.e. Give a new starting address to the array. Is it possible?
> >
> > Thanks in advance for all Your replies.
> >
> > A.
> >
> Search this group for numerous discussion about hazels of converting int
> to pointer.
>
> Standard has added special type intptr_t and uintptr_t to enable integer
> types to hold object pointers.
> Discussed in section:
> "7.18.1.4 Integer types capable of holding object pointers".
>
> But this is a *nice to have*, your compiler can decide not implement
> this data type.
>
> Again, this is only for the conversion from/to void.

I think no compiler has it implemented as yet ? Is this so?
>
>
> To answer your specific questions.
> Q1: Both are undefined behavior
>
> Q2: Both means the same i.e. address of the first element.
> [ However the moment you pass an array to a function, the meaning
> changes, and hence meaning of array and &array differs there ]

Can you be more elaborate on this please?

>
> Q3&Q4: array is not an lvalue, see section:
> 6.3.2.1 Lvalues, arrays, and function designators
> So you cannot modify it.
> ( Again the meaning of array changes the moment you pass it to
> function)
>
> --
> (Welcome) http://www.ungerhu.com/jxh/clc.welcome.txt
> (clc FAQ) http://www.eskimo.com/~scs/C-faq/top.html

Can I get a copy of the standard for download from somewhere?

Thanks in advance!

Suman wrote:
> pemo wrote:
>> "anonymous" <> wrote in message
>> news: ups.com...
> [snip]
>>> Question 2:
>>> What is the difference between array and &array. Are not they the same
>>> thing i.e. starting
>> an array name decomposes into the address of its first element, i.e., a
>> constant.
>>
> [snip]
>
> Well, almost.
>
>> You obviously cannot take the address of a 'value', and so &array
>> doesn't make sense, and I assume the compiler's saying to itself 'well, ok,
>> I know what he means'.
>
> In a value context, an object of type "array N of T" becomes
> a value of type "pointer to T", pointing to the first element
> of that array, i.e., the one with subscript 0.

Not when it is the operand of either sizeof or the unary &, then it does
*not* degenerate to a pointer. So the type of &array is pointer to
array, where as the type of pointer array degenerates to is pointer to
T. So they are different types.

> And do search the archives.

Also check the FAQ where it will be far easier to find the correct
information.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.

Krishanu Debnath wrote:
> anonymous wrote:
>> I have couple of questions related to array addresses. As they belong
>> to the same block, I
>> am putting them here in one single post. I hope nobody minds:
>>
>> char array[35];
>>
>> int address;

<snip>

>> Question 2:
>> What is the difference between array and &array. Are not they the same
>> thing i.e. starting
>> address of the array.
>
> array and &array are same in value context, both yield pointer to
> first element in value context.

No, they have different types. &array has the type pointer to array of
35 char, where as array degenerates to a pointer to char.

Check the comp.lang.c FAQ

<snip>
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.

anonymous wrote:
> Anand wrote:
>> anonymous wrote:
>>> I have couple of questions related to array addresses. As they belong
>>> to the same block, I
>>> am putting them here in one single post. I hope nobody minds:
>>>
>>> char array[35];
>>>
>>> int address;
>>>
>>> Questions 1:
>>> Why cannot I do the following:
>>>
>>> address = (int)array;
>>>
>>> whereas I found it perfectly alright to do the following:
>>>
>>> address = (int)&array;
>>>
>>> Question 2:
>>> What is the difference between array and &array. Are not they the same
>>> thing i.e. starting
>>> address of the array.

<snip>

>> Q2: Both means the same i.e. address of the first element.
>> [ However the moment you pass an array to a function, the meaning
>> changes, and hence meaning of array and &array differs there ]
>
> Can you be more elaborate on this please?

It's not quite right. &array is always different from array, they differ
in type. So, given your definition above, the compiler *must* complain
if you pass &array to a function with a prototype in scope specifying a
pointer to char. E.g.

#include <string.h>
int main(void)
{
char array[35] = "hello";
size_t len1 = strlen(array); /* this is allowed */
size_t len2 = strlen(&array); /* The compiler is required to complain */
}

&array has type pointer to array, where as array otherwise degenerates
to a pointer to char.

>> Q3&Q4: array is not an lvalue, see section:
>> 6.3.2.1 Lvalues, arrays, and function designators
>> So you cannot modify it.
>> ( Again the meaning of array changes the moment you pass it to
>> function)
>>
>> --
>> (Welcome) http://www.ungerhu.com/jxh/clc.welcome.txt
>> (clc FAQ) http://www.eskimo.com/~scs/C-faq/top.html
>
> Can I get a copy of the standard for download from somewhere?

Google for n1124 and you will find the draft for C99 + all the
subsequent corrections. That and all the other publicly available copies
are at http://www.open-std.org/jtc1/sc22/wg14/www/docs/ or you can buy a
copy of the actual release version.

I think though that you really need to work through a good text book
though, such as K&R2, and also read the FAQ (the URL is above) which
will tell you what K&R2 is (in the Bibliography, I think).
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.