@padnabhan.. u r correct..its clear now..thanks for the idea. On Tue, Jul 27, 2010 at 4:35 AM, Varma <[email protected]> wrote:
> > 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 <[email protected]> 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 < > > > > > > > > [email protected]> 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.. :) > > > > > -- > > > You received this message because you are subscribed to the Google > Groups > > > "Algorithm Geeks" group. > > > To post to this group, send email to [email protected]. > > > To unsubscribe from this group, send email to > > > [email protected]<algogeeks%[email protected]> > <algogeeks%2bunsubscr...@googlegroups .com> > > > . > > > For more options, visit this group at > > >http://groups.google.com/group/algogeeks?hl=en. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]<algogeeks%[email protected]> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
