array to pointer decay question !!

[pandapower]

Hi,
I know about the equivalence of pointer and arrays.But my doubt
comes when its for multidimentiona l arrays.I have read the C faq but
still have some doubts.

Suppose I have a declaration as
1. char array[x][x] or char array[][x]
and it will get decayed to
2. char *array[x]

similarly if I have a declaration as
3. char *argv[]
it can decay to
4. char **argv
Though its at our prerogative the way we use it.


Now from the above cases we see that,
1. 2 dimentional array can get decayed to an array of char pointers.
2. An array of char pointers can also be used as a pointer to pointer.

If I am correct till now, what I mean to know is that though we have
the cases above,a 2 dimentional array can never directly decay to a
pointer to pointer.


Correct me if I am wrong.

regards
pandapower

[Joona I Palaste]

Re: array to pointer decay question !!

pandapower <pandapower@sof thome.net> scribbled the following:[color=blue]
> Hi,
> I know about the equivalence of pointer and arrays.But my doubt
> comes when its for multidimentiona l arrays.I have read the C faq but
> still have some doubts.[/color]
[color=blue]
> Suppose I have a declaration as
> 1. char array[x][x] or char array[][x]
> and it will get decayed to
> 2. char *array[x][/color]

Yes.
[color=blue]
> similarly if I have a declaration as
> 3. char *argv[]
> it can decay to
> 4. char **argv
> Though its at our prerogative the way we use it.[/color]

Yes.
[color=blue]
> Now from the above cases we see that,
> 1. 2 dimentional array can get decayed to an array of char pointers.
> 2. An array of char pointers can also be used as a pointer to pointer.[/color]
[color=blue]
> If I am correct till now, what I mean to know is that though we have
> the cases above,a 2 dimentional array can never directly decay to a
> pointer to pointer.[/color]

That is true, i.e. it can never directly decay into a pointer to a
pointer. The rule is, more or less:
An array of type <T> can decay into a pointer to type <T>, but only
once. This new, decayed type, can never further decay into anything.

--
/-- Joona Palaste (palaste@cc.hel sinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"'I' is the most beautiful word in the world."
- John Nordberg

[Peter Nilsson]

Re: array to pointer decay question !!

"pandapower " <pandapower@Sof tHome.net> wrote in message
news:3713912d.0 402020156.3ec7d 4bf@posting.goo gle.com...[color=blue]
> Hi,
> I know about the equivalence of pointer and arrays.[/color]

They are not equivalent! ;)
[color=blue]
> But my doubt comes when its for multidimentiona l arrays.
> I have read the C faq but still have some doubts.[/color]

Try Steve Summit's course notes as additional reading...

http://www.eskimo.com/~scs/cclass/notes/sx10.html

[color=blue]
>
> Suppose I have a declaration as
> 1. char array[x][x] or char array[][x]
> and it will get decayed to
> 2. char *array[x][/color]

If and what an array 'decays to' depends on the context, but 2 is wrong.

char array[2];
char *p = array; /* array decays to &array[0] */

....

char array[2][5];
char (*p)[5] = array; /* array decays to &array[0]
// which is a pointer to an
// array of 5 char
*/

size_t size = sizeof array; /* array doesn't decay */
char (*q)[2][5] = &array; /* array doesn't decay */

The unary & and sizeof operators are exempt, when array is the only operand.
[color=blue]
> similarly if I have a declaration as
> 3. char *argv[]
> it can decay to
> 4. char **argv
> Though its at our prerogative the way we use it.
>
> Now from the above cases we see that,
> 1. 2 dimentional array can get decayed to an array of char pointers.[/color]


Now you appear to be talking about function parameters, not 'ordinary'
objects. In such cases, arrays don't 'decay' as a such. It's just that there
happen to be two different syntax methods for the _same_ thing...

int main(int argc, char *argv[]);
int main(int argc, char **argv);

void foo(int array[20][50]);
void foo(int array[][50]);
void foo(int (*array)[50]);

void bar(double array[20][50][7]);
void bar(double array[][50][7]);
void bar(double (*array)[50][7]);

C does not pass arrays by value (i.e. a copy of the whole array is not
passed), it only passes pointers to sequences of objects. So, the array
syntax in function parameter declarations are actually pointer declarations.
[color=blue]
> 2. An array of char pointers can also be used as a pointer to pointer.[/color]

Vaguely yes. A named object which is an array decays to a pointer.
[color=blue]
> If I am correct till now, what I mean to know is that though we have
> the cases above,a 2 dimentional array can never directly decay to a
> pointer to pointer.[/color]

Correct. A two dimensional array can only decay to a pointer to an array.

--
Peter

[Martin Dickopp]

Re: array to pointer decay question !!

pandapower@Soft Home.net (pandapower) writes:
[color=blue]
> I know about the equivalence of pointer and arrays.[/color]

There is no such equivalence. The relation between pointers and arrays
is that if an array is used in a value context, it is converted to a
pointer to its first element.
[color=blue]
> Suppose I have a declaration as
> 1. char array[x][x] or char array[][x]
> and it will get decayed to
> 2. char *array[x][/color]

Yes, if by "decay to" you mean "is converted to in a value context."
[color=blue]
> similarly if I have a declaration as
> 3. char *argv[]
> it can decay to
> 4. char **argv[/color]

Well, I don't think saying it *can* decay is correct. In a value
context, it is *always* converted.

Also, note the difference between the types of `a' and `b' in these
examples:

void foo (void) { char a [42]; }

void bar (char b []) { }

`a' has type array-of-char, and gets converted to a pointer in a value
context. `b' already has type pointer-to-char, because there's special
rule for parameter declarations. In a parameter declaration (and only
there!), the innermost [] operator is just a fancy way to declare a
pointer.
[color=blue]
> Now from the above cases we see that,
> 1. 2 dimentional array can get decayed to an array of char pointers.
> 2. An array of char pointers can also be used as a pointer to pointer.
>
> If I am correct till now, what I mean to know is that though we have
> the cases above,a 2 dimentional array can never directly decay to a
> pointer to pointer.[/color]

Correct. All that happens is that `array of T' is converted to `pointer
to T' in a value context. This is also true if `T' is itself an array,
so `array of array of U' is converted to `pointer to array of U'.

Martin

[Arjan Kenter]

Re: array to pointer decay question !!



Joona I Palaste wrote:[color=blue]
> pandapower <pandapower@sof thome.net> scribbled the following:
>[color=green]
>>Hi,
>> I know about the equivalence of pointer and arrays.But my doubt
>>comes when its for multidimentiona l arrays.I have read the C faq but
>>still have some doubts.[/color]
>
>[color=green]
>>Suppose I have a declaration as
>>1. char array[x][x] or char array[][x]
>>and it will get decayed to
>>2. char *array[x][/color]
>
>
> Yes.[/color]

No, it decays to char (*array)[x], which is very different:
it's a pointer to an array of chars, while your 2nd declaration
is an array of pointers to char.
[color=blue][color=green]
>>similarly if I have a declaration as
>>3. char *argv[]
>>it can decay to
>>4. char **argv
>>Though its at our prerogative the way we use it.[/color]
>
>
> Yes.
>
>[color=green]
>>Now from the above cases we see that,
>>1. 2 dimentional array can get decayed to an array of char pointers.[/color][/color]

No, it decays to a pointer to array to char.
[color=blue][color=green]
>>2. An array of char pointers can also be used as a pointer to pointer.[/color][/color]

Yes, indeed.
[color=blue][color=green]
>>If I am correct till now, what I mean to know is that though we have
>>the cases above,a 2 dimentional array can never directly decay to a
>>pointer to pointer.[/color]
>
>
> That is true, i.e. it can never directly decay into a pointer to a
> pointer. The rule is, more or less:
> An array of type <T> can decay into a pointer to type <T>, but only
> once. This new, decayed type, can never further decay into anything.[/color]

Unless you repeat the trick:

int x[10][42];
int *y = (x + 1)[2];

x + 1 has type: int (*x)[42] (decayed once)
(x + 1)[2] has type: int * (decayed again)

But of course our wonderful FAQ will make all this much clearer.

--
ir. H.J.H.N. Kenter ^^
Electronic Design & Tools oo ) Philips Research Labs
Building WAY 3.23 =x= \ arjan.kenter@ph ilips.com
Prof. Holstlaan 4 (WAY31) | \ tel. +31 40 27 45334
5656 AA Eindhoven /|__ \ tfx. +31 40 27 44626
The Netherlands (____)_/ http://www.kenter.demon.nl/

Famous last words: Segmentation Fault (core dumped)

[Horst Kraemer]

Re: array to pointer decay question !!

On 2 Feb 2004 01:56:12 -0800, pandapower@Soft Home.net (pandapower)
wrote:
[color=blue]
> Hi,
> I know about the equivalence of pointer and arrays.[/color]

There isn't such a thing. The correct formulation would be: If used in
an expression (except in the expressions sizeof array or &array) an
array of <T> decays to (is implicitly converted to) a pointer to <T>
which points to the initial array element.

But my doubt[color=blue]
> comes when its for multidimentiona l arrays.I have read the C faq but
> still have some doubts.
>
> Suppose I have a declaration as
> 1. char array[x][x] or char array[][x]
> and it will get decayed to
> 2. char *array[x][/color]

No. The expression 'array' will decay to a pointer which may be
produced by the declaration

char (*array) [x]

i.e. pointer to array of x chars. Your definition

char *array[x]

declares a array of x pointers to char which is a very different
thing.

char array[2][5]

is by definition an array of 2 (arrays of 5 chars). Read from left to
right. The first, i.e. only the top 'array' decays to 'pointer'. Same
for

char array [2][3][4];

array of 2 (arrays of 3 (arrays of 4 chars))

This decays to

pointer to (array of (3 arrays of 4 chars)), i.e. to

char (*array) [3][4]

and not to

char * array [3][4]

( array of 3 (arrays of 4 pointers to char) )


--
Horst