Re: [Dev-C++] whats wrong with this?

[Jonathan W. <jon...@gm...>]

You're welcome,

If you're still confused by references, consider this: they are the same as
pointers, i.e. they reference other data, but C++ hides some of the dangers
of pointers with them: you cannot change the pointer that the reference
uses, plus they simplify notation where you don't need the extra
liberty/functionality.

>I was aware of the leak [...]
I suspected that :)

Regards,
Jonathan

On 5/21/07, Lloyd <ll...@cd...> wrote:
>
> Thank you Jonathan for your vary detailed reply. I works now. (I was
> confused with the concept of call by reference :(  ). I was aware of the
> leak, but just to demonstrate the problem I wrote this program.
>
> Thank you very much for your quick reply.
>
> Thanks and Regards,
> Lloyd
>
>
>
> On Mon, 2007-05-21 at 11:14 +0200, Jonathan Winterflood wrote:
> > Hi,
> >
> > You are missing a level of indirection:
> > the 'a' you are setting here (a=temp) is a local variable in the
> > function 'call', so you can access the  int that it points to, but not
> > the pointer that main knows
> >
> > The a that main knows is always an uninitialized pointer, and you're
> > just lucky it's not segfaulting :D
> >
> > to change the pointer main knows, you need a pointer or a reference to
> > that pointer. ie.:
> > (by pointer)
> > bool call(int **a) {
> >  static int x;
> >  ++x;
> >  int *temp=new int;
> >  *temp=x;
> >  *a=temp;
> >  cout<<"call->"<<x<<","<<**a<<endl;
> >  return  true;
> > }
> > and
> >  call(&a);
> >
> > or
> > (by reference)
> > bool call(int* &a) { // no further changes
> >
> > Also, you have a memory leak: you loose the new ints you create
> > (indeed it's a small leak, but consider this: I was leaking 192bytes
> > in a loop and in 2 minutes, I ate up all my 2Gb of RAM....)
> >
> > Regards,
> > Jonathan
> >
> > On 5/21/07, Lloyd <ll...@cd...> wrote:
> >         Hi ,
> >
> >         This is a small program and its output
> >
> >         #include<iostream>
> >         using namespace std;
> >
> >         bool call(int *a)
> >         {
> >         static int x;
> >         ++x;
> >         int *temp=new int;
> >         *temp=x;
> >         a=temp;
> >         cout<<"call->"<<x<<","<<*a<<endl;
> >         return  true;
> >         }
> >
> >         int main()
> >         {
> >         int *a;
> >         for(int i=0;i<10;++i)
> >         {
> >           call(a);
> >           cout<<"val->"<<*a<<endl;
> >         }
> >         return 0;
> >         }
> >
> >         Outputcall->1,1
> >         val->0
> >         call->2,2
> >         val->0
> >         call->3,3
> >         val->0
> >         ......
> >         ...
> >
> >         The output which I wanted is
> >
> >         Outputcall->1,1
> >         val->1
> >         call->2,2
> >         val->2
> >         call->3,3
> >         val->3
> >         ......
> >         ....
> >
> >         The modification of the pointer in the function is reflected
> >         in the
> >         calling function. Would anybody tell me the reason for this
> >         mistake?
> >
> >         Thanks and Regards,
> >         Lloyd
> >
> >
> >
> >
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