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]

Re: Why multiplication not allowed?

"Vijay Kumar R Zanvar" <vijoeyz@hotpop .com> writes:
[color=blue]
> Why multiplication of pointers is not allowed?
> Till now I only know this, but not the reason why![/color]

What would multiplication of pointers mean?

[Vijay Kumar R Zanvar]

Re: Why multiplication not allowed?


"Ben Pfaff" <blp@cs.stanfor d.edu> wrote in message
news:87ekudv9da .fsf@pfaff.stan ford.edu...[color=blue]
> "Vijay Kumar R Zanvar" <vijoeyz@hotpop .com> writes:
>[color=green]
> > Why multiplication of pointers is not allowed?
> > Till now I only know this, but not the reason why![/color]
>
> What would multiplication of pointers mean?[/color]

#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]

Re: Why multiplication not allowed?

"Vijay Kumar R Zanvar" <vijoeyz@hotpop .com> writes:
[color=blue]
> "Ben Pfaff" <blp@cs.stanfor d.edu> wrote in message
> news:87ekudv9da .fsf@pfaff.stan ford.edu...[color=green]
> > "Vijay Kumar R Zanvar" <vijoeyz@hotpop .com> writes:
> >[color=darkred]
> > > Why multiplication of pointers is not allowed?
> > > Till now I only know this, but not the reason why![/color]
> >
> > What would multiplication of pointers mean?[/color]
>
> #include <stdio.h>
> #include <stdlib.h>
>
> int
> main ( void )
> {
> int i = 10;
> int *k = &i, *j = &i;
>
> k = k * j;[/color]

You mean
*k = *k * *j;
In other words, multiply `int's, not pointers to `int's.
[color=blue]
> return EXIT_SUCCESS;
> }[/color]

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]

Re: Why multiplication not allowed?


"Ben Pfaff" <blp@cs.stanfor d.edu> wrote in message
news:877k05v8c2 .fsf@pfaff.stan ford.edu...[color=blue][color=green]
> > int i = 10;
> > int *k = &i, *j = &i;
> >
> > k = k * j;[/color]
>
> You mean
> *k = *k * *j;
> In other words, multiply `int's, not pointers to `int's.
>[/color]

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]

Re: Why multiplication not allowed?

Vijay Kumar R Zanvar wrote:[color=blue]
> "Ben Pfaff" <blp@cs.stanfor d.edu> wrote in message
> news:87ekudv9da .fsf@pfaff.stan ford.edu...
>[color=green]
>>"Vijay Kumar R Zanvar" <vijoeyz@hotpop .com> writes:
>>
>>[color=darkred]
>>> Why multiplication of pointers is not allowed?
>>>Till now I only know this, but not the reason why![/color]
>>
>>What would multiplication of pointers mean?[/color]
>[/color]

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]

Re: Why multiplication not allowed?

"Vijay Kumar R Zanvar" <vijoeyz@hotpop .com> wrote:
[color=blue]
> Why multiplication of pointers is not allowed?[/color]

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]

Re: Why multiplication not allowed?

Julian Maisano wrote:
[color=blue]
>
> "Ben Pfaff" <blp@cs.stanfor d.edu> wrote in message
> news:877k05v8c2 .fsf@pfaff.stan ford.edu...[color=green][color=darkred]
>> > int i = 10;
>> > int *k = &i, *j = &i;
>> >
>> > k = k * j;[/color]
>>
>> You mean
>> *k = *k * *j;
>> In other words, multiply `int's, not pointers to `int's.
>>[/color]
>
> Is the following code wrong? I mean, does it do the same?
> int* k =&i
> int* j=&i
> k* = k* * j*[/color]

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.
[color=blue]
> (It seems to me more readable than the above. I come from Pascal
> background)[/color]

In C, the dereferencing operator comes before the pointer. C programmers are
accustomed to this syntax, and find it perfectly readable.


--
Richard Heathfield : binary@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]

Re: Why multiplication not allowed?

Richard Heathfield wrote:[color=blue]
> Julian Maisano wrote:[color=green]
>> Is the following code wrong? I mean, does it do the same?
>> int* k =&i
>> int* j=&i
>> k* = k* * j*[/color]
>
> 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 *[/color]

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

i* = j;

is also a syntax error.

Jeremy.

[Richard Heathfield]

Re: Why multiplication not allowed?

Jeremy Yallop wrote:
[color=blue]
> Richard Heathfield wrote:[color=green]
>> Julian Maisano wrote:[color=darkred]
>>> Is the following code wrong? I mean, does it do the same?
>>> int* k =&i
>>> int* j=&i
>>> k* = k* * j*[/color]
>>
>> 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 *[/color]
>
> It doesn't parse that way. The whitespace between '*' and '=' is
> significant,[/color]

Indeed it is. Oops, sorry, darn and heck. (More or less in that order.)
[color=blue]
> so
>
> i* = j;
>
> is also a syntax error.[/color]

Right.

--
Richard Heathfield : binary@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

[Eric Sosman]

Re: Why multiplication not allowed?

Vijay Kumar R Zanvar wrote:[color=blue]
>
> 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.[/color]

The result would be meaningless. An easy (but
informal) way to understand this is to use an analogy:
think of pointer values as street addresses. Then
the following operations make sense:

- Find the house at "123 Main Street."

- To find the neigboring house, compute "123
Main Street plus one." (I'm ignoring such
real-world intrusions as odd-even numbering
schemes, discontinuities between city blocks,
and so on: we're just exploring an imperfect
analogy, after all.)

- To find the distance between two Main Street
addresses, compute "123 Main Street minus 189
Main Street," yielding "minus 66 lots."

However, some other arithmetical operations make
no sense at all:

- Computing "123 Main Street plus 207 Main Street"
produces no useful answer.

- Computing "123 Main Street minus 123 Elm Street"
produces no useful answer.

- Similarly, computing "123 Main Street times
89 El Camino Real" makes no sense. Perhaps the
result might be considered a kind of "area,"
but there seems to be no useful analogous
concept to "area" in the addressing of memory-
resident objects.

- Finally, computing "123 Main Street times three"
might possibly make sense, arriving at "369 Main
Street." But such a definition carries a built-
in assumption that Main Street is zero-based,
and in a linear computer memory no more than one
object can begin at "address zero." If you want
to find the house three times as far from the
start of the street, you really want to compute
"Main Street Origin plus three times (123 Main
Street minus Main Street Origin)."

--
Eric.Sosman@sun .com

[Sean Kenwrick]

Re: Why multiplication not allowed?


"Ben Pfaff" <blp@cs.stanfor d.edu> wrote in message
news:877k05v8c2 .fsf@pfaff.stan ford.edu...[color=blue]
> "Vijay Kumar R Zanvar" <vijoeyz@hotpop .com> writes:
>[color=green]
> > "Ben Pfaff" <blp@cs.stanfor d.edu> wrote in message
> > news:87ekudv9da .fsf@pfaff.stan ford.edu...[color=darkred]
> > > "Vijay Kumar R Zanvar" <vijoeyz@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?[/color]
> >
> > #include <stdio.h>
> > #include <stdlib.h>
> >
> > int
> > main ( void )
> > {
> > int i = 10;
> > int *k = &i, *j = &i;
> >
> > k = k * j;[/color]
>
> You mean
> *k = *k * *j;
> In other words, multiply `int's, not pointers to `int's.
>[/color]

Your solution above is correct but can you figure out why the following code
is invalid (which it is):


*k = *k/*j;

assuming k and j are pointers to ints that have been correctly
initialized....

Sean

[Jarno A Wuolijoki]

Re: Why multiplication not allowed?

On Tue, 6 Jan 2004, Sean Kenwrick wrote:
[color=blue]
> Your solution above is correct but can you figure out why the following code
> is invalid (which it is):
> *k = *k/*j;[/color]

*k = k*//*j;

[Jarno A Wuolijoki]

Re: Why multiplication not allowed?

On Tue, 6 Jan 2004, Jarno A Wuolijoki wrote:
[color=blue][color=green]
> > Your solution above is correct but can you figure out why the following code
> > is invalid (which it is):
> > *k = *k/*j;[/color]
>
> *k = k*//*j;[/color]

Should have engaged my brain. It starts a new one, duh.

[Ben Pfaff]

Re: Why multiplication not allowed?

"Sean Kenwrick" <skenwrick@hotm ail.com> writes:
[color=blue]
> Your solution above is correct but can you figure out why the following code
> is invalid (which it is):
>
>
> *k = *k/*j;[/color]

There's no way to tell whether it's invalid unless we can see
what follows it.
--
"Welcome to the wonderful world of undefined behavior, where the demons
are nasal and the DeathStation users are nervous." --Daniel Fox