Purvi mehta

• Nov 8th, 2006

 

Ans:- No,&arr will store the address of array.

yuvraj

• Mar 13th, 2007

 

The answer is YES bcoz both arr and &arr gives the base address of an array.
You can try it by a simple example
void main()
{
 int arr[10];
 printf("%dn",arr);
 printf("%d",&arr);
}

Jyotiranjan Moahnty

• Oct 8th, 2007

 

#include<stdio.h>
int main()
{
 int arr[2];
 printf("%dn",arr);
 printf("%dn",(arr+1));
 printf("%dn",&arr);
 printf("%dn",(&arr+1));
}
 

Try it, I think it is the better example for you to understand that, arr and &arr is not same. Their starting address is same, but the value in that memory is not same.

I think you are a student. I appreciate, your involvement with this  kind of things.

indirakannan26

 Profile Answers by indirakannan26 

• Aug 12th, 2008

 

Sorry, it is not same because arr stores the value where as &arr is the address of that value.

KishoreKNP

 Profile Answers by KishoreKNP 

• Dec 4th, 2009

 

Arr and &Arr both are same, which will return the address of first element of the array. For example

void main()
{
   char arr[10] = "Kishore";
   printf("n%u : %u", a, &a);
}
Output:
~~~~~

65516 : 65516

Mohamed

• May 8th, 2013

 

Actually arr and &arr are different.
arr holds the base address of the array arr[], but &arr holds the address of the entire array.
but both expressions points to the same base address.

Code
  #include<stdio.h>


int main()


{


 int arr[2]={5,6};


 printf("%u


",arr);


 printf("%u


",(arr+1));


 printf("%u


",&arr);


 printf("%u


",(&arr+1));


return 0;


}


 

ruphaensil

• Dec 31st, 2013

 

error...because a is undefined