EXERCISES ON POINTERS

READING 
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Kernighan and Ritchie, 5.1-5.3
King, Chapter 11 (Pointers)
      Chapter 12.1-12.3 (Pointers and arrays)

SUPPLEMENTAL READING
--------------------
Deitel and Deitel, 7.1-7.4 7.7-7.8


EXERCISES
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 0. What's the difference between the following declarations.

         A.  int *x;
         B.  int* x;
         C.  int * x;

 1. Assume x = 5, y = 10.  What are the values of x and y after
    calling swap1(x, y)?

        void swap1(int a, int b) {
           int t;
           t = a; a = b; b = t;
        }

 2. Assume x = 5, y = 10.  What are the values of x and y after
    calling swap2(&x, &y)?

        void swap2(int *pa, int *pb) {
           int t;
           t = *pa; *pa = *pb; *pb = t;
        }


 3. Assume int x[5] = {0, 1, 2, 3, 4}.  What are the values of
    x after calling swap3(x, 1, 4)?

        void swap3(int a[], int i, int j) {
           int t;
           t = a[i]; a[i] = a[j]; a[j] = t;
        }

 4. Assume int x[5] = {0, 1, 2, 3, 4}.  What are the values of
    x after calling swap2(x+1, x+4)?

 5. Assume int x[5] = {0, 1, 2, 3, 4}.  What does print3(&x[0])
    print? print3(&x[2])? print3(&x[4])?

        void print3(int x[]) {
           int i;
           for (i = 0; i < 3; i++)
              printf("%d ", x[i]);
        }

 6. When we pass an array to a function in C, it passes a pointer
    to the 0th array element instead.  Why doesn't C just create
    a new local copy of the array, as it does with integers?

 7. What is the difference between the following two functions

       int middle1(int a[], int n) {      int middle2(int *a, int n) {
          return a[n/2];                     return a[n/2];
       }                                  }

 8. To print an integer n to standard output we use printf("%d", n),
    but to read it in we use scanf("%d ", &n). Why is the & necessary
    with scanf?
    

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		    ANSWERS TO EXERCISES ON POINTERS

 0. None. Some people prefer A to indicate that *x is an int;
    thus x is a pointer to an int. Others prefer B to indicate that
    x is the variable being declared. Not many people use C; it was
    just included to demonstrate that the position of the * is 
    irrelevant.
   

 1. x = 5, y = 10. The function has no effect since integers are
    passed by value.  Variables a and b contain only local copies
    of the integers 5 and 10.

 2. x = 10, y = 5. Since the memory addresses of x and y are passed,
    the function is able to correctly swap their values.

 3. x = {0, 4, 2, 3, 1}.  The memory address of the 0th array
    element of a is passed to the function.  

 4. x = {0, 4, 2, 3, 1}.  Address arithmetic says that a+1 is the
    memory address of the 1st array element and a+4 is a pointer to
    the 4th.

 5. 0 1 2  Same as print3(a).
    2 3 4  Same as print3(a+2). Just like array starts at element 2.
    4 ? ?  Out of bounds access.

 6. Efficiency.  Arrays can have billions of elements.  The overhead
    of copying them each time you call a function would be overwhelming.
    It also means that you can pass a "subarray" to a function for
    processing as in the previous question.

 7. None. It is a matter of style. Beginners usually prefer the first.

 8. scanf needs to change the value of n.  A function cannot change the
    value of an integer unless it knows its memory address.  Instead
    of passing the current value of n, we use &n to pass its memory address.