dev-cpp-users

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

[Lloyd <ll...@cd...>]

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




______________________________________
Scanned and protected by Email scanner

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

[Lloyd <ll...@cd...>]

Sorry some typing error are occured... please check this mail...

On Mon, 2007-05-21 at 14:25 +0530, Lloyd 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;
> }
> 
> Output

> call->1,1
> val->0
> call->2,2
> val->0
> call->3,3
> val->0
> .......
> ....
> 
> The output which I wanted is 
> 
> Output

> call->1,1
> val->1
> call->2,2
> val->2
> call->3,3
> val->3
> ......
> .....
> 
> The modification of the pointer in the function is not reflected in the
> calling function. Would anybody tell me the reason for this mistake?
> 
> Thanks and Regards,
>  Lloyd
> 
> 
> 
> 
> ______________________________________
> Scanned and protected by Email scanner
> 
> -------------------------------------------------------------------------
> This SF.net email is sponsored by DB2 Express
> Download DB2 Express C - the FREE version of DB2 express and take
> control of your XML. No limits. Just data. Click to get it now.
> http://sourceforge.net/powerbar/db2/
> _______________________________________________
> Dev-cpp-users mailing list
> Dev...@li...
> TO UNSUBSCRIBE: http://www23.brinkster.com/noicys/devcpp/ub.htm
> https://lists.sourceforge.net/lists/listinfo/dev-cpp-users


______________________________________
Scanned and protected by Email scanner

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
>
>
>
>
> ______________________________________
> Scanned and protected by Email scanner
>
> -------------------------------------------------------------------------
> This SF.net email is sponsored by DB2 Express
> Download DB2 Express C - the FREE version of DB2 express and take
> control of your XML. No limits. Just data. Click to get it now.
> http://sourceforge.net/powerbar/db2/
> _______________________________________________
> Dev-cpp-users mailing list
> Dev...@li...
> TO UNSUBSCRIBE: http://www23.brinkster.com/noicys/devcpp/ub.htm
> https://lists.sourceforge.net/lists/listinfo/dev-cpp-users
>



-- 
<Morpheus> linux, c'est une question de VI ou de MORE

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

[Lloyd <ll...@cd...>]

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
>         
>         
>         
>         
>         ______________________________________ 
>         Scanned and protected by Email scanner
>         
>         -------------------------------------------------------------------------
>         This SF.net email is sponsored by DB2 Express
>         Download DB2 Express C - the FREE version of DB2 express and
>         take 
>         control of your XML. No limits. Just data. Click to get it
>         now.
>         http://sourceforge.net/powerbar/db2/
>         _______________________________________________
>         Dev-cpp-users mailing list 
>         Dev...@li...
>         TO UNSUBSCRIBE:
>         http://www23.brinkster.com/noicys/devcpp/ub.htm
>         https://lists.sourceforge.net/lists/listinfo/dev-cpp-users
> 
> 
> 
> -- 
> <Morpheus> linux, c'est une question de VI ou de MORE


______________________________________
Scanned and protected by Email scanner

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
> >
> >
> >
> >
> >         ______________________________________
> >         Scanned and protected by Email scanner
> >
> >
> -------------------------------------------------------------------------
> >         This SF.net email is sponsored by DB2 Express
> >         Download DB2 Express C - the FREE version of DB2 express and
> >         take
> >         control of your XML. No limits. Just data. Click to get it
> >         now.
> >         http://sourceforge.net/powerbar/db2/
> >         _______________________________________________
> >         Dev-cpp-users mailing list
> >         Dev...@li...
> >         TO UNSUBSCRIBE:
> >         http://www23.brinkster.com/noicys/devcpp/ub.htm
> >         https://lists.sourceforge.net/lists/listinfo/dev-cpp-users
> >
> >
> >
> > --
> > <Morpheus> linux, c'est une question de VI ou de MORE
>
>
> ______________________________________
> Scanned and protected by Email scanner
>



-- 
<Morpheus> linux, c'est une question de VI ou de MORE

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

[Per W. <pw...@ia...>]

The a pointer in the function call() is just a copy of the a pointer in
main(). Hence, your allocated pointer will not be transmitted back.

You must modify call() to either take a reference to a pointer, or send i
a pointer to a pointer. This will give call() the address of the pointer a
in main(), so that it knows where to store the pointer value returned from
new.

/pwm

On Mon, 21 May 2007, Lloyd 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
>
>
>
>
> ______________________________________
> Scanned and protected by Email scanner
>
> -------------------------------------------------------------------------
> This SF.net email is sponsored by DB2 Express
> Download DB2 Express C - the FREE version of DB2 express and take
> control of your XML. No limits. Just data. Click to get it now.
> http://sourceforge.net/powerbar/db2/
> _______________________________________________
> Dev-cpp-users mailing list
> Dev...@li...
> TO UNSUBSCRIBE: http://www23.brinkster.com/noicys/devcpp/ub.htm
> https://lists.sourceforge.net/lists/listinfo/dev-cpp-users
>