Mathematics 1201: Programming in C
		Solutions to Homework Assignment #6
			Prof. Wickerhauser

Exercise 1, Chapter 6, p. 138.
  The function and a program to test it are shown below:
    #include <stdio.h>
    void Swap( int *p1, int *p2) {
      int tmp;
      tmp = *p1;      *p1 = *p2;      *p2 = tmp;
      return;
    }
    void Sort3( int *A, int *B, int *C ) {
      if(*A>*B) Swap( A, B );	/* now A<=B */
      if(*A>*C) Swap( A, C );	/* now A<=C and A<=B */
      if(*B>*C) Swap( B, C );	/* now B<=C */
      return;
    }
    main(void) {
      int a, b, c;
      a=1; b=2; c=3; Sort3(&a, &b, &c); printf("a,b,c=%d,%d,%d\n",a,b,c);
      a=2; b=1; c=3; Sort3(&a, &b, &c); printf("a,b,c=%d,%d,%d\n",a,b,c);
      a=2; b=3; c=1; Sort3(&a, &b, &c); printf("a,b,c=%d,%d,%d\n",a,b,c);
      a=3; b=2; c=1; Sort3(&a, &b, &c); printf("a,b,c=%d,%d,%d\n",a,b,c);
      a=3; b=1; c=2; Sort3(&a, &b, &c); printf("a,b,c=%d,%d,%d\n",a,b,c);
      a=1; b=3; c=2; Sort3(&a, &b, &c); printf("a,b,c=%d,%d,%d\n",a,b,c);
      return 0;
    }

Exercise 8, Chapter 6, p. 139.
    (a) The address of A, which is machine-dependent
    (b) The value dereferenced by Ptr, namely the value of A, which is 5
    (c) False, namely 0
    (d) True, namely nonzero
    (e) The address of Ptr, which is machine-dependent
    (f) The content of memory location 5, which is machine-dependent and
	    may be invalid
    (g) The value of A, namely 5
    (h) The value dereferenced by Ptr, namely the value of A, which is 5

Extra Problem 1:
	main(void) {
	  const int * x;
	  int const * y;
	  int * const z;
	  x = 0;
	  y = 0;
	  z = 0;   <----- first warning: trying to assign a constant
	  *x = 1;   <----- second warning: trying to assign a constant
	  *y = 2;   <----- third warning: trying to assign a constant
	  *z = 3;
	  return 0;
	}
  If the program compiles, then at run time it will crash at the line
	*x = 1;
  since x is 0 and *0 is guaranteed to be an invalid dereference.

Extra Problem 2:
(a)	% a.out
	xp1 = 67107140, xp2 = 67107144, xp2-xp1 = 1
(b) 4 memory locations, the difference between xp1 and xp2, are used to store
	a float.  The value of xp2-xp1 is the number of floats that can be
	stored in the space between xp1 and xp2.