pointers

[darklight]

Which is the correct way, to initialise a pointer to an element of
an array. I am sorry if my terminology is not correct

/* declare pointer */
int *ptr;
/* define and initialise array */
int array[10] = {,0,1,2,3,4,5,6,7,8,9};
/* assign value to pointer */
ptr = &array[9]; or ptr = (array + 9);

I have found both work but to avoid future problems is one
better than the other. And if so why

[Richard Heathfield]

darklight wrote:

Which is the correct way, to initialise a pointer to an element of
an array. I am sorry if my terminology is not correct

/* declare pointer */
int *ptr;
/* define and initialise array */
int array[10] = {,0,1,2,3,4,5,6,7,8,9};
/* assign value to pointer */
ptr = &array[9]; or ptr = (array + 9);

I have found both work but to avoid future problems is one
better than the other. And if so why

The two expressions are equivalent. Each points ptr to the last element of
the array.

In a value context, the following pairs of expressions are equivalent:

(arrayname) == (&arrayname[0])
(arrayname + n) == (&arrayname[n])
(*(arrayname + n)) == (arrayname[n])

Which you choose is entirely a matter of taste.

I should perhaps add that the external parentheses are there purely as an
attempted defence against nitpicking! :-)

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Al Bowers]

darklight wrote:

Which is the correct way, to initialise a pointer to an element of
an array. I am sorry if my terminology is not correct

/* declare pointer */
int *ptr;
/* define and initialise array */
int array[10] = {,0,1,2,3,4,5,6,7,8,9};
/* assign value to pointer */
ptr = &array[9]; or ptr = (array + 9);

I have found both work but to avoid future problems is one
better than the other. And if so why

Either way will be valid. Generally you will find the [] operator
the most accepted method. To me, it it easier to understand the
operations.

By the way the () operator is not required in the above assignment,
but you may add them if it gives you more clarity. It could be,
ptr = array + 9;

However, in dereferencing, you will need the () operator to yield
the value of the last element of the array named array.
int i = *(array + 9;) will assign to i the value 9.
int i = *array + 9; will assign to i the value 18.

--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapidsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

[Derk Gwen]

darklight <ng******@netscapel.net> wrote:
# Which is the correct way, to initialise a pointer to an element of
# an array. I am sorry if my terminology is not correct
#
# /* declare pointer */
# int *ptr;
# /* define and initialise array */
# int array[10] = {,0,1,2,3,4,5,6,7,8,9};
# /* assign value to pointer */
# ptr = &array[9]; or ptr = (array + 9);
#
# I have found both work but to avoid future problems is one
# better than the other. And if so why

&(array[9]) = &(*(array+9)) = (array+9)

The notations are equivalent, and you can expect a compiler to generate
the exact same code for both. Which is better is style question for you.

Which do you think looks better?
Which do you more easily understand?

That is the correct one.

--
Derk Gwen http://derkgwen.250free.com/html/index.html
Don't say anything. Especially you.

[Eric]

darklight <ng******@netscapel.net> wrote:

Which is the correct way, to initialise a pointer to an element of
an array. I am sorry if my terminology is not correct

/* declare pointer */
int *ptr;
/* define and initialise array */
int array[10] = {,0,1,2,3,4,5,6,7,8,9};
/* assign value to pointer */
ptr = &array[9]; or ptr = (array + 9);

I have found both work but to avoid future problems is one
better than the other. And if so why

Either one works equally well and the decision on which to use is
generally based on personal preference and the context.
--
== Eric Gorr ========= http://www.ericgorr.net ========= ICQ:9293199 ===
"Therefore the considerations of the intelligent always include both
benefit and harm." - Sun Tzu
== Insults, like violence, are the last refuge of the incompetent... ===

[Anand]

Hi

ptr=&array[9] 0r (array+9) both refer same operation. The basic difference
is that by accessing with pointers the operation will be quite faster while
comparing array[9]. Hence good C programmers will prefer pointers. Hope this
information will be useful to u. If u find anymore idea like this please let me
know.

Regards,
Anand.

[pete]

Anand wrote:

Hi

ptr=&array[9] 0r (array+9) both refer same operation.
The basic difference is that by accessing with pointers
the operation will be quite faster while comparing array[9].

Unless it isn't faster.

The compiler has enough information to know that
(&array[9]) and (array+9) are the same.

I'm unfamiliar with the address operator as being translated
into a time consuming machine operation.

--
pete

[Horst Kraemer]

On 12 Jan 2004 02:53:40 -0800, an********@yahoo.co.in (Anand) wrote:

Hi

ptr=&array[9] 0r (array+9) both refer same operation. The basic difference
is that by accessing with pointers the operation will be quite faster while
comparing array[9]. Hence good C programmers will prefer pointers. Hope this
information will be useful to u. If u find anymore idea like this please let me
know.

My compilers generate the same code for

p = &array[9]

and

p = array+9

What about your compilers?

Regards
Horst

[Richard Heathfield]

Anand wrote:

Hi

ptr=&array[9] 0r (array+9) both refer same operation. The basic
difference
is that by accessing with pointers the operation will be quite faster
while comparing array[9].
Get A Better Compiler. Yours Is Broken.

Hence good C programmers will prefer pointers.
Better C programmers know that the compiler knows perfectly well that the
two expressions are equivalent.
Hope this information will be useful to u.

Let us hope that u, whoever he is, never reads it, since it's fundamentally
flawed.

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Christopher Benson-Manica]

Eric <eg*************@verizon.net> spoke thus:

darklight <ng******@netscapel.net> wrote:

/* declare pointer */
int *ptr;
/* define and initialise array */
int array[10] = {0,1,2,3,4,5,6,7,8,9}; ^ extra comma removed /* assign value to pointer */
ptr = &array[9]; or ptr = (array + 9);

Either one works equally well and the decision on which to use is
generally based on personal preference and the context.

But the equivalence breaks down for

ptr=array+10; /* Legal */
ptr=&array[10]; /* Illegal (?), since array[10] is a dereference */

, right?

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

[Jeremy Yallop]

Christopher Benson-Manica wrote:

Eric <eg*************@verizon.net> spoke thus:

darklight <ng******@netscapel.net> wrote:

/* declare pointer */
int *ptr;
/* define and initialise array */
int array[10] = {0,1,2,3,4,5,6,7,8,9}; ^ extra comma removed /* assign value to pointer */
ptr = &array[9]; or ptr = (array + 9);

Either one works equally well and the decision on which to use is
generally based on personal preference and the context.

But the equivalence breaks down for

ptr=array+10; /* Legal */
ptr=&array[10]; /* Illegal (?), since array[10] is a dereference */

, right?

Wrong, at least in C99.

[6.5.3.2]
3 The unary & operator returns the address of its operand. [...]
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.

Jeremy.

[Christopher Benson-Manica]

Jeremy Yallop <je****@jdyallop.freeserve.co.uk> spoke thus:

3 The unary & operator returns the address of its operand. [...]
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.

Thank you, I am yet again abashed yet wiser :)

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

[Clark Cox]

In article <cd************************@posting.google.com>,
an********@yahoo.co.in (Anand) wrote:

Hi

ptr=&array[9] 0r (array+9) both refer same operation.
You should have just stopped there.
The basic difference
is that by accessing with pointers the operation will be quite faster while
comparing array[9].

Says who? I have *never* seen a compiler that generates more efficent
code for (&array[9]) than it does for *(array+9).