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%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]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
