Re: Better way of allocating a memory( auto_ptr vs new vs malloc)

=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <>
Sun, 07 Dec 2008 09:34:17 GMT
On 2008-12-06 23:36, blargg wrote:

Ami wrote:

My god that is so hard to read. Look at the difference a couple of
find-and-replaces and some indention do:

And it lets us find some things to improve upon.

I have few qustions
Consider the allocation

  T p(T());

  T p(); will do just as well.

T() in above would do default initialization right
then I can safely avoid the use of C style memset..
and then nevere have to de allocate the same.

As far as auto_ptr is concrned , do I again need to initialize to
T () as shown below

  auto_ptr<T>p = (new T()) ;

No need to use () around the new, auto_ptr<T>p = new T(); works just as

another major concerned abt using auto_ptr is

after the memory allocation is done tht is lets say

  int main ()
      auto_ptr<T>p = (new T()) ;
      func(p.get()); //
      return ( EXIT_SUCESS);
  void func(T *output)
    // some code

do u see any problem with uisng an auto pointer like above i.e passing
as an argument in a function call .

No problem at all, but you might want to consider using a shared pointer
and pass that instead of using get(). Or pass a reference to the auto ptr.

Is there any memory Leak ?

No, but there is always the risk that func() will somehow save the
pointer you pass in and someone else will later try to access it.

I think the auto of auto ptr is the best way ..
Also in the code the malloc is done like

  T *p = (T *)malloc (sizeof(T)*(n+1));

I'm not sure how it is in C++, but in C you do not have to cast the
result of malloc.

so using new for above is like

  T *p = (T *)new char [sizeof(T)*(n+1)];

  T* p = new T[n+1];

Erik Wikstr??m

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