This is the code given in my algorithms book.We need to calculate its space complexity.
this is the answer given-
The space complexity is S=C+Sp and in this case Sp is zero as the code is independent of n.
But I wanted to calculate the C part of the code and in this case it would be 5*2bytes=10 bytes considering a,x,n,0 and -1.
So my question is what would be the C in case a was a 2d matrix,do we consider it as 2 bytes only as in the case of 1d array or we take it as 4 bytes?
Pointers and integers are no more 16 bits since the mid-eighties so I wonder what you mean with your 2 bytes.
Also note that your accounting of space is a little weird. You don't count the variable i, but you do count literal constants, which usually appear as immediate arguments in the assembly code.
Anyway, on a 32 bits machine, all addresses (and pointers) are represented on 4 bytes and the extra accounting of array dimensions is performed by the compiler and hard-coded into the assembly.
On a 64 bits machine, count 8 bytes per pointer and 4 bytes per int.