Re: [Dev-C++] RES: function for allocate memory for 2D vectors

As you can see, the code become simler when you return the allocated
array.

Two things:
If you fail to allocate a column, it might be a good idea to clean up any
partial allocations before returning NULL. This is important if you intend
to continue to run the program after a failed allocation. A 10k x 10x
array of 32-bit integers would consume about 400MB, so an allocation
failure could potentially leak 400MB memory.

Then rename Alocate2DVect to Allocate2DVect and everything should be ok :)

/pwm

EMA System        email: pw...@ia...
Per Westermark    web: http://iapetus.neab.net
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Sweden


On Sun, 11 Nov 2007, Andr=E9 Mac=E1rio Barros wrote:

> My way of understand your recommendations
> did result in this:
>
> //////code///////////
> #include <stdio.h>
> #include <stdlib.h>
>
> int **Alocate2DVect(int lin, int col)
> {
>    int i, **vect;
>
>    vect =3D (int **) malloc(lin*sizeof(int));
>    if(!vect) return(NULL);
>    for(i=3D0; i<col; i++){
>       vect[i] =3D (int *) malloc(col*sizeof(int));
>       if(!vect[i]) return(NULL);
>    }
>    return(vect);
> }
>
> int main()
> {
>    int **vector2D;
>    int i, j, lin=3D3, col=3D4;
>
>    vector2D =3D Alocate2DVect(lin, col);
>    if(!vector2D) exit(EXIT_FAILURE);
>
>    for(i=3D0; i<lin; i++)
>       for(j=3D0; j<col; j++)
>          vector2D[i][j] =3D i + j;
>
>    for(i=3D0; i<lin; i++){
>       for(j=3D0; j<col; j++){
>          printf("%d\t", vector2D[i][j]);
>       }
>       printf("\n");
>    }
>
>    for(j=3D0; j<col; j++)
>       free(vector2D[col]);
>    free(vector2D);
>
>    system("PAUSE");
>    return 0;
> }
>
> ///////// end of code ////////
>
> Is it ok?
>
> Thanks!
> Andr=E9
>
>
> -----Mensagem original-----
> De: Per Westermark [mailto:pw...@ia...]
> Enviada em: domingo, 11 de novembro de 2007 04:27
> Para: Andr=E9 Mac=E1rio Barros
> Cc: dev...@li...
> Assunto: Re: [Dev-C++] function for allocate memory for 2D vectors
>
>
> First of all - your code would be a bit simpler (you would loose one
> indirection) if you modify Allocate2DVect (two l in allocate) to return
> the pointer instead of taking it as a parameter.
>
> Second: your test if (!vect) does not check the result of the allocation,
> but if you sent a pointer in to the function.
>
> Third: You normally don't need to release memory before ending a program,
> but if you do, then your code is incorrect. You need to release all
> columns before releasing the row vector.
>
> /pwm
>
> On Sun, 11 Nov 2007, Andr=E9 Mac=E1rio Barros wrote:
>
> > Dear Users,
> >
> >    Could you confirm if this function is
> > correct?
> >    The reason of my question is: inspecting
> > the address of *vect in line x in the function
> > is different of line y. I was expecting equal
> > addresses, wasn=B4t?
> >
> > ////////// code /////////////////////////
> >
> > int Alocate2DVect(int ***vect, int lin, int col)
> > {
> >    int i, *pont;
> >
> >    *vect =3D (int **) malloc(lin*sizeof(int)); // line x
> >    if(!vect) return(0);
> >    for(i=3D0; i<col; i++) *(*vect + i) =3D (int *) malloc(col*sizeof(in=
t));
> >    if(!vect) return(0); // line y
> >    return(1);
> > }
> >
> =09> int main()
> > {
> >    int **vector2D;
> >    int i, j, lin=3D3, col=3D4;
> >
> >    if(!Alocate2DVect(&vector2D, lin, col)) exit(1);
> >
> >    for(i=3D0; i<lin; i++)
> >       for(j=3D0; j<col; j++)
> >          vector2D[i][j] =3D i + j;
> >
> >    for(i=3D0; i<lin; i++){
> >       for(j=3D0; j<col; j++){
> >          printf("%d\t", vector2D[i][j]);
> >       }
> >       printf("\n");
> >    }
> >
> >    free(vector2D);
> >    system("PAUSE");
> >    return 0;
> > }
> > //////////// end of code /////////
> >
> > Regards
> > Andre
> >
> >
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> 10/11/2007
> > 15:47
> >
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