In declaration of function parameters, int a[5] is equivalent to int *a.
A
Correct answer
Explanation
In C function parameter declarations, array notation always decays to a pointer. When you write 'int a[5]' as a function parameter, it is automatically converted to 'int *a' by the compiler. The array size (5 in this case) is ignored - the compiler only cares that it's a pointer to an integer. This is why array parameters don't carry size information.