Question

/* * Return 1 if ptr points to an element within the specified intArray, 0 otherwise. * Pointing anywhere in the array is fair game, ptr does not have to * point to the beginning of an element. Check the spec for examples if you are * confused about what this method is determining. * size is the size of intArray in number of ints. Can assume size != 0. * Operators / and % and loops are NOT allowed. *

* ALLOWED: * Pointer operators: *, & * Binary integer operators: -, +, *, <<, >>, ==, ^ * Unary integer operators: !, ~ * Shorthand operators based on the above: ex. <<=, *=, ++, --, etc. *

* DISALLOWED: * Pointer operators: [] (Array Indexing Operator) * Binary integer operators: &, &&, |, ||, <, >, !=, /, % * Unary integer operators: - */ int withinArray(int * intArray, int size, int * ptr) {\

Also not allowed to use constructs like if, while, for, etc.

Answer 1

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Answer 2

int withinArray(int * intArray, int size, int * ptr) {

      if(ptr == NULL) // if ptr == NULL return 0

            return 0;

      // if end of intArr is reached

      if(size == 0)

          return 0; // element not found

  else  

          {

            if(*(intArray+size-1) == *(ptr)) // check if (size-1)th element is equal to ptr

                   return 1; // return 1

            return withinArray(intArray, size-1,ptr); // recursively call the function with size-1 elements

         }

}