Code problem

Keith Thompson wrote:
> Angel<angel+news@spamcop.net> writes:
> [...]
>> Anyway, the bottom line to the original poster is "don't mix up integers
[....]

Only scenario I could think of is on a 68k System where
one woud want to stick a function pointer into Address
4, generate a Trap and hope memory wasn't corrupted :-P


--
See why I hate Windows users?
All pain, no gain.
-Howard Chu

On Fri, 8 Jun 2012 07:52:18 -0700 (PDT), rage <hansum.rahul@gmail.com>
wrote:

>Can someone please help me out on this?
>
>#include<stdio.h>
>main()
>{
> int i;
> int *j=10;
> i=j+19;

j is an address...


*j is a value



> printf("%d %d\n",j,i);

i = *j + 10 = 29

i = j + 10 = address + 10 == garbage

>}
>
>The output is a garbage value (86) on my pc and ideone. Should it not be 29?


IMHO

MJR

"rage" <hansum.rahul@gmail.com> wrote in message
news:7081cbd2-1ff6-4591-89c0-e7680b330790@googlegroups.com...
> Can someone please help me out on this?
>
> #include<stdio.h>
> main()
> {
> int i;
> int *j=10;
> i=j+19;
> printf("%d %d\n",j,i);
> }
>
> The output is a garbage value (86) on my pc and ideone. Should it not be
> 29?

As others have correctly pointed out, you have violated a dozen or so
constraints.

Using "old style" C (prior to the ISO standards), here is the answer I would
give:

If you add an integer to a pointer, the integer is *scaled* by the size of
the object that the pointer points to. An "int" on your machine must be
four bytes, therefore 19 is multiplied by 4 before adding it to the 10.
Thus you get 86. (4*19+10)

One of the important things to remember here: arithmetic involving pointers
is *scaled*.

--

numerist at aquaporin4 dot com