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

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

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(int)=
);
>    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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