assigning array addresses to integer variables and vice versa

[anonymous]

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.

[pemo]

Re: assigning array addresses to integer variables and vice versa


"anonymous" <call_ret@yahoo .com> wrote in message
news:1132833621 .437259.58740@z 14g2000cwz.goog legroups.com...[color=blue]
>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.[/color]
[color=blue]
>Questions 1:
>Why cannot I do the following:[/color]
[color=blue]
>address = (int)array;[/color]

What do you by 'can't do'?

[color=blue]
> Question 2:
> What is the difference between array and &array. Are not they the same
> thing i.e. starting[/color]

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'.

[color=blue]
> Question 3:
>
> Why cannot I do array++. Is array a pointer constant?[/color]

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

[color=blue]
> Question 4:
> How can I do the following:
>
> array = address;[/color]


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

[Krishanu Debnath]

Re: assigning array addresses to integer variables and vice versa

anonymous wrote:[color=blue]
> 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;[/color]

No, it is a legal syntax. However pointer to integer conversion may
invoke undefined behavior if result cannot be represented in integer
type.
[color=blue]
>
> 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.[/color]

array and &array are same in value context, both yield pointer to
first element in value context.
[color=blue]
>
> Question 3:
>
> Why cannot I do array++. Is array a pointer constant?[/color]

postfix increment operator needs a modifiable l-value. array name is
not a modifiable l-value.
[color=blue]
>
> Question 4:
> How can I do the following:
>
> array = address;
>
> i.e. Give a new starting address to the array. Is it possible?[/color]

No, see above.

Krishanu

[anonymous]

Re: assigning array addresses to integer variables and vice versa


pemo wrote:[color=blue]
> "anonymous" <call_ret@yahoo .com> wrote in message
> news:1132833621 .437259.58740@z 14g2000cwz.goog legroups.com...[color=green]
> >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;
> >[/color][/color]
[color=blue][color=green]
> > 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.[/color]
>[color=green]
> >Questions 1:
> >Why cannot I do the following:[/color]
>[color=green]
> >address = (int)array;[/color]
>
> What do you by 'can't do'?[/color]

Ooops, checking again, it is perfectly alright to do
address = (int)array;
My fault somewhere.[color=blue]
>[color=green]
> > Question 2:
> > What is the difference between array and &array. Are not they the same
> > thing i.e. starting[/color]
>
> 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'.[/color]

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

Thanks for your help.

[Anand]

Re: assigning array addresses to integer variables and vice versa

anonymous wrote:[color=blue]
> 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.
>[/color]
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

[Suman]

Re: assigning array addresses to integer variables and vice versa


pemo wrote:[color=blue]
> "anonymous" <call_ret@yahoo .com> wrote in message
> news:1132833621 .437259.58740@z 14g2000cwz.goog legroups.com...[/color]
[snip][color=blue][color=green]
> > Question 2:
> > What is the difference between array and &array. Are not they the same
> > thing i.e. starting[/color]
>
> an array name decomposes into the address of its first element, i.e., a
> constant.
>[/color]
[snip]

Well, almost.
[color=blue]
> 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'.[/color]

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.

[anonymous]

Re: assigning array addresses to integer variables and vice versa


Anand wrote:[color=blue]
> anonymous wrote:[color=green]
> > 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.
> >[/color]
> 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.[/color]

I think no compiler has it implemented as yet ? Is this so?[color=blue]
>
>
> 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 ][/color]

Can you be more elaborate on this please?
[color=blue]
>
> 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[/color]

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

Thanks in advance!

[Eric Sosman]

Re: assigning array addresses to integer variables and vice versa

anonymous wrote:
[color=blue]
> 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;[/color]

You can, if you like. The result will be useful on
some machines, useless on others. I wouldn't recommend it.
[color=blue]
> whereas I found it perfectly alright to do the following:
>
> address = (int)&array;[/color]

That's equally legitimate, equally useful or useless.
Again, I wouldn't recommend it.
[color=blue]
> Question 2:
> What is the difference between array and &array. Are not they the same
> thing i.e. starting
> address of the array.[/color]

This is Question 6.12 in the comp.lang.c Frequently Asked
Questions (FAQ) list

http://www.eskimo.com/~scs/C-faq/top.html

[color=blue]
> Question 3:
>
> Why cannot I do array++. Is array a pointer constant?[/color]

FAQ Question 6.7.
[color=blue]
> Question 4:
> How can I do the following:
>
> array = address;
>
> i.e. Give a new starting address to the array. Is it possible?[/color]

No.
[color=blue]
> Thanks in advance for all Your replies.[/color]

You're welcome. Read the FAQ; it will do you good.

--
Eric Sosman
esosman@acm-dot-org.invalid

[Eric Sosman]

Re: assigning array addresses to integer variables and vice versa

Krishanu Debnath wrote:
[color=blue]
> array and &array are same in value context, both yield pointer to
> first element in value context.[/color]

They are not the same. Both produce pointers, but
point to different objects. Please read the FAQ.

--
Eric Sosman
esosman@acm-dot-org.invalid

[Jordan Abel]

Re: assigning array addresses to integer variables and vice versa

On 2005-11-24, Eric Sosman <esosman@acm-dot-org.invalid> wrote:[color=blue]
> Krishanu Debnath wrote:
>[color=green]
>> array and &array are same in value context, both yield pointer to
>> first element in value context.[/color]
>
> They are not the same. Both produce pointers, but
> point to different objects. Please read the FAQ.[/color]

They will compare equal, though, if converted to a void *

[because they point to the same memory location]

[Flash Gordon]

Re: assigning array addresses to integer variables and vice versa

Suman wrote:[color=blue]
> pemo wrote:[color=green]
>> "anonymous" <call_ret@yahoo .com> wrote in message
>> news:1132833621 .437259.58740@z 14g2000cwz.goog legroups.com...[/color]
> [snip][color=green][color=darkred]
>>> Question 2:
>>> What is the difference between array and &array. Are not they the same
>>> thing i.e. starting[/color]
>> an array name decomposes into the address of its first element, i.e., a
>> constant.
>>[/color]
> [snip]
>
> Well, almost.
>[color=green]
>> 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'.[/color]
>
> 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.[/color]

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.
[color=blue]
> And do search the archives.[/color]

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.

[Flash Gordon]

Re: assigning array addresses to integer variables and vice versa

Krishanu Debnath wrote:[color=blue]
> anonymous wrote:[color=green]
>> 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;[/color][/color]

<snip>
[color=blue][color=green]
>> Question 2:
>> What is the difference between array and &array. Are not they the same
>> thing i.e. starting
>> address of the array.[/color]
>
> array and &array are same in value context, both yield pointer to
> first element in value context.[/color]

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.

[Flash Gordon]

Re: assigning array addresses to integer variables and vice versa

anonymous wrote:[color=blue]
> Anand wrote:[color=green]
>> anonymous wrote:[color=darkred]
>>> 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.[/color][/color][/color]

<snip>
[color=blue][color=green]
>> 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 ][/color]
>
> Can you be more elaborate on this please?[/color]

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.
[color=blue][color=green]
>> 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[/color]
>
> Can I get a copy of the standard for download from somewhere?[/color]

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.

[Keith Thompson]

Re: assigning array addresses to integer variables and vice versa

Anand <Anand@no-replies.com> writes:[color=blue]
> anonymous wrote:[color=green]
>> 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.
>>[/color]
> 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[/color]

Converting a pointer to an integer *can* invoke undefined behavior.
Specifically:

Any pointer type may be converted to an integer type. Except as
previously specified, the result is implementation-defined. If the
result cannot be represented in the integer type, the behavior is
undefined. The result need not be in the range of values of any
integer type.
[color=blue]
> 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 ][/color]

As others have pointed out, they're not the same; they have different
types.

There's nothing special about function calls. An array expression is
implicitly converted to a pointer to its first element in *any*
context other than as the operand of a unary "&" or "sizeof", or when
it's a string literal in an initializer.

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Jordan Abel]

Re: assigning array addresses to integer variables and vice versa

On 2005-11-24, Keith Thompson <kst-u@mib.org> wrote:[color=blue][color=green]
>> 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 ][/color]
>
> As others have pointed out, they're not the same; they have different
> types.
>
> There's nothing special about function calls. An array expression is
> implicitly converted to a pointer to its first element in *any*
> context other than as the operand of a unary "&" or "sizeof", or when
> it's a string literal in an initializer.[/color]

well, there _is_ one thing special about functions, but it pertains to
function prototypes, not funtion calls: the fact that declaring what
appears to be an array formal parameter is in fact a pointer formal
parameter [and thus, int main(int argc, char *argv[5]); is compatible
with int main(int argc, char **argv);, and sizeof argv == sizeof(char
**) in both cases, and not 5*sizeof(char *).]