Answer / pooja sonawane

the index of array, which is of the form a[i], is converted
by the compiler in the form [a+i]. So, the index of first
element is zero because [a+0] will give 'a' & the first
array element can be accessed. Due to this, we can also
access the array elements as'i[a]' instead of 'a[i]' , &
this will not produce an error.

Answer / chandan singh from niit

This boils down to the concept of Binary digits. Take an
array size of 64 for example. We start from 0 and end at
63. We require 6 bits.But, if we were to start from 1 and
end at 64, we would require 7 bits to store the same
number, thus increasing the storage size.

Answer / rck

The complier treats a[0] as *(a+0) ie *(a+0) (array name is
the base address of that array)gives the first element of
the array.
If the array starts from 1 a[1] is treated as *(a+1) gives
the second element of array.The first element is missing.so
array always starts from 0.

Answer / ashish verma

The index of array, which is of the form a[i], is converted
by the compiler in the form [a+i]. So, the index of first
element is zero because [a+0] will give 'a' & the first
array element can be accessed. Due to this, we can also
access the array elements as'i[a]' instead of 'a[i]' , &
this will not produce an error.

Answer / ranga raj.b

Array is a chunk of continues memory allocation. The arthmetic that is used by compiler to move to the next array location is StartAddress + (Index)*sizeof(int)if it is an integer array.now if we see to access the first location of the array we have to give Index =0 and for next location index =1 and so on...This is why array index start with zero.

Answer / chetan

Both Answer ARE correct but First one is EXACT and after
that compiler is design according to that

Answer / ashlesha

Second answer is correct. but its compiler specific to start an array index with 1 or zero.as in pascal the array index starts with '1'.

Answer / vikas chand

yes first ans is exactly correct.
second one is wrong because in java, c# etc array index is
start with zero..
but in that case we can not write a[i] as i[a]...
because a is reference variable...
so i think first ans is exactly right......

Answer / sabab ali khan

becouse the value of given index are multiplied by size of
datatype of array and then add in the base Address to find
address of that block which you want to access
ex.
int arr[5];
if the base address of array is 1001 then last block
address will be 1009
and if we give arr[2] then compiler calculate address like
this..
1001 + sizeof(int)* 2 = 1005
where base address = 1001
and 2 is the given index.
so the calculated address is 1005 and third block will be
access.

Answer / v .saikiran

I Think second one correct

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