RE: [Dev-C++] Number of elements of a multidimensional

[Ioannis V. <no...@ho...>]

> -----Original Message-----
> From: dev...@li...=20
> [mailto:dev...@li...] On Behalf=20
> Of Carlos Garc=EDa del Monte
> Sent: Thursday, November 28, 2002 6:54 PM
> To: dev-cpp-users-list
> Subject: Re: [Dev-C++] Number of elements of a multidimensional
>=20
>=20
> The sizeof applied to a array will give you the size of a=20
> pointer, never
> the size of the array.
> Say char *array; or char array[];

As a function argument, because only then the second is a pointer.
Inside the scope of its definition, a

char array[]=3D"12345678901234567890";


size_t x=3Dsizeof(array);


x will always be 21  (the '\0' included).



However in void f(char array[]);


is equivallent to void f(char *);


so you can do array++; for example because all tthese are ponters.


You can do that even in the case:


void f(char array[20])
{
 /* The !=3D'\0' is not needed */
 while(*array!=3D'\0')
   *array++=3D'a';
}

All "array" arguments of functions are pointers but we can use the array
syntax for notational purposes.



> Then sizeof(array) =3D 4. (at least in my intelIII x86, with win2k 4 =
is
> the size for any kind of pointer).
>=20
> The only solution is to have some special value for the end of the
> array. Which for strings is '\0', and you can try to set another
> depending on which kind of data you're working with and which range of
> values you can find, so you can then choose on of the values which are
> out of the range to set as that special value. So you can reach for it
> and you can then now, with the approppriate functions you can=20
> write the
> total size of the array and the size of each row of the=20
> multidimensional
> array.


Yes he has to "remember" the size of the array when he passes it across
functions, even by passing the sizes as additional arguments, or other
tricks.


Ioannis Vranos
=20
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