Re: Allocating memory for an object twice

From:

Arne Mertz <news@arne-mertz.de>

Newsgroups:

comp.lang.c++

Date:

Wed, 25 Mar 2009 00:23:22 +0100

Message-ID:

<49c96b69$0$31866$9b4e6d93@newsspool3.arcor-online.net>

Victor Bazarov schrieb:

prasoonthegreat@gmail.com wrote:

#include<iostream>
using std::cout;
using std::endl;

const int size=3;

template<class T>
class vector
{
    T *v;

    public:

    vector()
    {
        v=new T[size];
        for(int i=0;i<size;i++)
           v[i]=0;

It's easier just to do

v=new T[size]();

(note the parentheses), and the array will be zero-initialised.

       cout<<"Hello 1";
    }

    vector(T *a)
    {
        //v=new T[size];
        for(int i=0;i<size;i++)
           v[i]=a[i];

What if the pointer points to an array that is smaller than 'size'? You
would get *undefined behaviour*!

Even IF a pointed to an array that is big enough this is the point where
the segfault occurs: we are in the constructor here, so since there is
no initialization list v might contain any unspecified value because it
is an uninitialized POD. So most likely you don't need to worry about
the size of the array a is pointing to because yet in the first
iteration the assignment tries to access the memory v is pointing to
which most likely is not yours so you get your segfault.

    cout<<"Hello";
    }

    /*void operator=(const T *a)
    {
       for(int i=0;i<size;i++)
           v[i]=a[i];
    }*/

    T operator*(const vector &y)
    {
        T prod=0;

        for(int i=0;i<size;i++)
           prod += this->v[i] * y.v[i];

        return prod;
    }
};

[..]

why m i getting seg fault

Read about "The Rule of Three".

V