Re: anything wrong with this code?

From:

John Harrison <john_andronicus@hotmail.com>

Newsgroups:

comp.lang.c++

Date:

Wed, 07 Mar 2007 21:27:32 GMT

Message-ID:

<87GHh.56854$z54.5588@newsfe3-gui.ntli.net>

hijkl wrote:

On Mar 7, 1:09 am, "hijkl" <s_mah...@yahoo.com> wrote:

On Mar 7, 12:07 am, John Harrison <john_androni...@hotmail.com> wrote:

hijkl wrote:

ok i tried running this program..it will run fine only problem with
this code is memory leak..
there atleast 12 int pointers that needs to free at the end.

thanks
sanjay

Only two int pointer that need to be freed.

Your main misunderstanding seem to be that you think this code

int *array(int n){
return new int(n);
}

allocates n integers.

HI JOhn
i understand that this code allocates only one intger.
but

        for( int i = 0; i < 10; i++ ) {
                p[i] = 0;
        }
this code will initialize next 9 integer to consecutive locations..
if you will debug then u will notice that..
let me know if you not agree with me.- Hide quoted text -

- Show quoted text -

John
try following code and you will understand how its working
...
for( int i = 0; i < 10; i++ ) {
p[i] = i;
}

printf("%d", [0]);
printf("%d", [1]);
printf("%d", [2]);
printf("%d", [3]);
.
.
.
printf("%d", [9]);
...
the output will be
0
1
2
.
.
.
9

so its cleary says that it allocated 10 integers.

thanks
sanjay

No it doesn't. You allocate 1 integer, and write the rest of your code
as if you have allocated 10. This is UNDEFINED BEHAVIOUR. How is what
your program doing different from undefined behaviour. Undefined
behaviour means ANYTHING CAN HAPPEN.

The rules of C++ say that new int(n) allocates one integer. That is the
only thing that matters.

john