sizeof(arr) is 4.. o.e the number of elements in the array
size of *arr is the size of any normal pointer i.e 4(in case of 32-bit
compilers)
so the answer is 1

On Sat, Jul 24, 2010 at 9:52 AM, ravi gupta <[email protected]> wrote:

>
>
> On Sat, Jul 24, 2010 at 9:40 AM, tarak mehta <[email protected]>wrote:
>
>> int arr[]={1,2,3,4};
>> k=sizeof(arr)/sizeof(*arr);
>> value of k=4;
>>
>> however
>>
>>
>> void hell(int arr[]);
>> main()
>> {
>>        int arr[]={1,2,3,4};
>>        hell(arr);
>> }
>> void hell(int arr[])
>> {
>> printf("%d",sizeof(arr)/sizeof(*arr));
>> }
>>
>>
>> output of hell() is 1. why???
>>
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>> When array is passed as an argument, only a pointer to the first element
> of the array is passed. Therefore the parameter int arr[] in void hell(int
> arr[]) is just a pointer, hence the result .
> I hope it answers your query.
>
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-- 
With Regards,
Jalaj Jaiswal
+919026283397
B.TECH IT
IIIT ALLAHABAD

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