@ monish : Beta monish welcum to algogeeks.....
they are not same ..
reason:
arr is the base address of array , i.e the address of first int
wheras &arr is the address of( array of ints) , they happen to be same but
they are nt..
nw consider the array
1 2 3 4 5
suppose array starting address is 6036
so
"arr" contains 6036 (as base address)
"&arr" contains 6036 (as address of array of ints)
now consider arr+ 1 --> this give 6036 + 4 =6040 (for 32 bit )
consider &arr+1 ---> this will give address of next array of 5 ints i .e
, 6056..
am i clear?
On Sat, Aug 13, 2011 at 4:59 PM, Dipankar Patro <dip10c...@gmail.com> wrote:
> [Quote]
>
> arr has the address to the base of an array of three int's.
>
> &arr is the address of the first element of that same array.
>
> [/Quote]
>
> ^^ Found on web.
>
> check the difference by printing arr+1 and &arr+1. The former will skip one
> element, but later will skip the whole array.
>
> On 12 August 2011 18:41, monish001 <monish.gup...@gmail.com> wrote:
>
>> Program:
>> int arr[] = {12, 14, 15, 23, 45};
>> printf("%u %u\n", arr, &arr);
>>
>> Question: Why arr == &arr ?
>>
>> Comments:
>> 1. arr is a variable that stores the address of location where arr[0]
>> resides. Complier shows &arr and arr having same value. Shouldn't &arr
>> be the address where arr resides?
>>
>> Thanks
>> Monish
>>
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