Question

For an array declared as int arr[50], calculate the address of arr[35], if Base(arr) =1000 and width=2​

Answer 1

Answer:

Explanation:

BASE(Arr)=1000 ,Width=2

address(35)=Base(add.)+w(35-L.B)

1000+2(35)=1070

address of arr[35] =1070

Answer 2

The address of arr[35] for the given array is $1070$.

Explanation:

• Let there be an array of 'n' elements having base address 'x'.
• Then the size of each element in memory address is referred as the width of the array. Let that be denoted by 'b'.
• Then the address 'A' of 'i' index of the array is given by, $A=x+bi\ \ \ \ \ \ \ \ \ \ \ \ \ \ ->0\leq i\leq (n-1)$  
• Here, we have an array 'arr' and $n=50,\ i=35,\ x=1000,\ b=2$.
• Then the required address is given by, $A_{35}=1000+2(35)\ \ \ \ \ \ \ \ \ \ ->A_{35}=1070$