Why multiplication not allowed?

[Vijay Kumar R Zanvar]

Hi,

Why multiplication of pointers is not allowed?
Till now I only know this, but not the reason why!

PS: As a rule, I searched the FAQ, but could not
find an answer.

--
Vijay Kumar R Zanvar
My Home Page - http://www.geocities.com/vijoeyz/

[Ben Pfaff]

"Vijay Kumar R Zanvar" <vi*****@hotpop .com> writes:

Why multiplication of pointers is not allowed?
Till now I only know this, but not the reason why!

What would multiplication of pointers mean?

[Vijay Kumar R Zanvar]

"Ben Pfaff" <bl*@cs.stanfor d.edu> wrote in message
news:87******** ****@pfaff.stan ford.edu...

"Vijay Kumar R Zanvar" <vi*****@hotpop .com> writes:

Why multiplication of pointers is not allowed?
Till now I only know this, but not the reason why!

What would multiplication of pointers mean?

#include <stdio.h>
#include <stdlib.h>

int
main ( void )
{
int i = 10;
int *k = &i, *j = &i;

k = k * j;
return EXIT_SUCCESS;
}

This gives an error:

ptr_mul.c: In function `main':
ptr_mul.c:10: error: invalid operands to binary *

[Ben Pfaff]

"Vijay Kumar R Zanvar" <vi*****@hotpop .com> writes:

"Ben Pfaff" <bl*@cs.stanfor d.edu> wrote in message
news:87******** ****@pfaff.stan ford.edu...

"Vijay Kumar R Zanvar" <vi*****@hotpop .com> writes:

Why multiplication of pointers is not allowed?
Till now I only know this, but not the reason why!
What would multiplication of pointers mean?

#include <stdio.h>
#include <stdlib.h>

int
main ( void )
{
int i = 10;
int *k = &i, *j = &i;

k = k * j;

You mean
*k = *k * *j;
In other words, multiply `int's, not pointers to `int's.
return EXIT_SUCCESS;
}

The difference between pointers and their referents is pretty
fundamental. Have you considered reading a textbook or taking a
class?
--
"I don't have C&V for that handy, but I've got Dan Pop."
--E. Gibbons

[Julian Maisano]

"Ben Pfaff" <bl*@cs.stanfor d.edu> wrote in message
news:87******** ****@pfaff.stan ford.edu...

int i = 10;
int *k = &i, *j = &i;

k = k * j;

You mean
*k = *k * *j;
In other words, multiply `int's, not pointers to `int's.

Is the following code wrong? I mean, does it do the same?
int* k =&i
int* j=&i
k* = k* * j*

(It seems to me more readable than the above. I come from Pascal background)

[Kevin Goodsell]

Vijay Kumar R Zanvar wrote:

"Ben Pfaff" <bl*@cs.stanfor d.edu> wrote in message
news:87******** ****@pfaff.stan ford.edu...

"Vijay Kumar R Zanvar" <vi*****@hotpop .com> writes:

Why multiplication of pointers is not allowed?
Till now I only know this, but not the reason why!

What would multiplication of pointers mean?

The question Ben is asking is, what would you propose pointer
multiplication do? How would you define the operation? I can't think of
any way in which multiplying a pointer by any other value would be
useful or even meaningful.

Your question is somewhat like asking "why can't I assign to a
function?" or "why can't I take the square root of a struct?" There's
just no logical reason why you *should* be able to.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Richard Bos]

"Vijay Kumar R Zanvar" <vi*****@hotpop .com> wrote:

Why multiplication of pointers is not allowed?

Jack lives at Westmoreland Street 12, London SW6 4E8. Jill lives at
Rossinistraat 27b, 1287 CZ Amsterdam. Please multiply these addresses
and let us know the result.

Richard

[Richard Heathfield]

Julian Maisano wrote:

"Ben Pfaff" <bl*@cs.stanfor d.edu> wrote in message
news:87******** ****@pfaff.stan ford.edu...

> int i = 10;
> int *k = &i, *j = &i;
>
> k = k * j;
You mean
*k = *k * *j;
In other words, multiply `int's, not pointers to `int's.

Is the following code wrong? I mean, does it do the same?
int* k =&i
int* j=&i
k* = k* * j*

Yes, it's wrong; no, it doesn't do the same thing; and no, it's not legal
syntax. The last line parses as:

k *= k * *j *

That is, multiply k (a pointer!) by the contents of j, multiply /that/ by
heaven knows what, and then use the result to multiply by k again,
assigning the result to k.
(It seems to me more readable than the above. I come from Pascal
background)

In C, the dereferencing operator comes before the pointer. C programmers are
accustomed to this syntax, and find it perfectly readable.
--
Richard Heathfield : bi****@eton.pow ernet.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

[Jeremy Yallop]

Richard Heathfield wrote:

Julian Maisano wrote:

Is the following code wrong? I mean, does it do the same?
int* k =&i
int* j=&i
k* = k* * j*

Yes, it's wrong; no, it doesn't do the same thing; and no, it's not legal
syntax. The last line parses as:

k *= k * *j *

It doesn't parse that way. The whitespace between '*' and '=' is
significant, so

i* = j;

is also a syntax error.

Jeremy.

[Richard Heathfield]

Jeremy Yallop wrote:

Richard Heathfield wrote:

Julian Maisano wrote:

Is the following code wrong? I mean, does it do the same?
int* k =&i
int* j=&i
k* = k* * j*
Yes, it's wrong; no, it doesn't do the same thing; and no, it's not legal
syntax. The last line parses as:

k *= k * *j *

It doesn't parse that way. The whitespace between '*' and '=' is
significant,

Indeed it is. Oops, sorry, darn and heck. (More or less in that order.)
so

i* = j;

is also a syntax error.

Right.

--
Richard Heathfield : bi****@eton.pow ernet.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