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

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

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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