1.Array name is nothing but a const pointer to the first element of an
array.
2.subscripting of an array can be visualized as incrementing a
pointer . Pointer always increments by the size of the type of data it
holds.
3. so, always sizeof(arr) is the sizeof a pointer , and sizeof(*arr)
is the sizeof(1) i.e size of the first element. it is an integer
here.
4. Pointers size and integers size are same in most compilers ( dont
think of far pointers now) . so it resulted in 1 despite of the no.of
elements in that array.
Hope i made it more clear :)
On Jul 25, 12:01 am, tarak mehta <tarakmeht...@gmail.com> wrote:
> void hell(int arr[]);
> main()
> {
> int arr[]={1,2,3,4,5};
> hell(arr);}
>
> void hell(int arr[])
> {
> printf("%d",sizeof(arr)/sizeof(*arr));}
>
> even this gives 1 !!
> @manjunath ur idea seems correct..but could u plz elaborate a bit
>
> On Sat, Jul 24, 2010 at 10:51 PM, Manjunath Manohar <
>
>
>
> manjunath.n...@gmail.com> wrote:
>
> > when arrays are passed as arguments to a function,the starting address of
> > the array is passed like a pointer,
> > thus sizeof(arr)=2..thus 2/2=1..this is the precise reason for always
> > specifying the column length in the definition of function when functions
> > have arrays as one of the arguments..
>
> > Hope i made any sense.. :)
>
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