NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/comp/lang/cplus/18305

Text File | 1992-12-22 | 4.5 KB | 189 lines

Newsgroups: comp.lang.c++
Path: sparky!uunet!cs.utexas.edu!qt.cs.utexas.edu!yale.edu!jvnc.net!newsserver.technet.sg!nuscc!papaya!suresh
From: [e-mail] (Suresh Thennarangam - Research Scholar)
Subject: Re: Can't I initialize a dynamically allocated object in its constructor
Message-ID: <[e-mail]>
Originator: suresh@papaya
Sender: [e-mail]
Organization: Institute of Systems Science, NUS, Singapore
Date: Tue, 22 Dec 1992 11:42:51 GMT
Lines: 177

In article  <[e-mail]>,[e-mail] (David Hansen)  writes:

In article <[e-mail]>, [e-mail] (C
ocotte Minute) writes:
> |> Hello,
> |>
> |> I have the following problem.
> |>
> |>  I have created an class: Vector(k), it's
> |> a vector of k dimension (of doubles). After a whole week of debugging,
> |> I have come to the conclusion that you must'nt set the values of the element


:
:
:
> The corrected code appears below.  Enjoy!
:
:
>
>
>
>
> #include <stdlib.h>    // exit
>
> //-------------------------------------------------------------------
> // Vector: vector of double
> //-------------------------------------------------------------------
>
> class Vector{
>
> double     *item ;
> int     size ;
>
>
> public :
>
> // constructors
> Vector() { item = new double[1] ; size = 1 ;}    // NOTE: use array form
> Vector(int array_size);
> Vector(const Vector & v);    // NOTE: new copy constructor
>
> // destructors
> ~Vector() { delete [] item ; }    // NOTE: use array form
>
> // methods
> int getsize() { return size ; }
>
> void print(char *message);
>
>
> // operators
> Vector operator+ (Vector& b);
> Vector& operator= (Vector& a);
> double& operator[] (int i);
> };
>
>
> // constructors
>
> Vector::Vector(int array_size) {
>    item = new double[array_size] ;
>    size = array_size ;
>    printf("constructing\n");
>    for(int i=0;i<array_size;i++)item[i]=-1.0;
> }
>
> // Definition of new copy constructor
> Vector::Vector(const Vector& v) {
>    int i;
>    size = v.size;
>    item = new double[size];
>    for (i=0; i<size; i++)
>       item[i] = v.item[i];
> }
>
>
>
>
> Vector& Vector::operator= (Vector& a){
>    if (getsize() != a.getsize()){
>       printf("operator '=':Vectors are not of the same size.\n");
>       exit(0);
>    }
>    for (int i = 0 ; i < a.getsize() ; i++){
>       item[i] = a.item[i] ;
>    }
>    return *this ;
> }
>
>
> double& Vector::operator[](int i){
>    return item[i] ;
> }
>
>
> void  Vector::print(char *message) {
>    printf("%s\n",message);
>    for (int i = 0 ; i < size ; i++){
>       printf("       [%d]=%lf\n", i, item[i]);
>    }
> }
>
>
>
> main(){
>
> Vector v1(3),v2(3),v3(3),v4(3);
>
>
>
> v1[0]=1.0;
> v1[1]=1.0;
> v1[2]=1.0;
>
> v2[0]=2.0;
> v2[1]=2.0;
> v2[2]=2.0;
>
> v3[0]=3.0;
> v3[1]=3.0;
> v3[2]=3.0;
>
> v4=v1+v2+v3;
>
> v4.print("v4");
>
> }
>
>
> Vector Vector::operator+ (Vector& a){
>    if (getsize() != a.getsize()){
>       printf("operator '+':Vectors are not of the same size.\n");
>       exit(0);
>    }
>    Vector retVect(getsize()) ;
>    for (int i = 0 ; i < a.getsize() ; i++){
>       retVect.item[i] = item[i] + a.item[i] ;
>    }
>    return retVect ;
> }
>
My apologies for including  the entire code above ....
Also to make things easier I have moved the function
Vector Vector::operator+() to the bottom.

My point:

retVect used above is a automatic variable that will go out of scope
as soon as the function Vector::operator+() is exited and it's
destructor will be called which will deallocate the memory
that was allocated by the constructor. Although the returned
Vector object is a copy and will have a valid item pointer, it will
be pointing to deallocated data.


Nitpicking:

The operator= function should check for self-assignment
like so   " if ( this != &a) ". Saves a lot of work



      __
     (_   / /  o_   o  o |_
     __)/(_( __) (_(_ /_)| )_



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* Suresh Thennarangam               *  EMail: [e-mail](Internet) *
* Research Scholar                  *         [e-mail]          *
* Institute Of Systems Science      *  Tel:  [phone].              *
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