Mathematics 1201: Programming in C
	       Solutions to Reading and Homework Assignment #2
			     Prof. Wickerhauser

Chapter 2 Exercises:

Exercise 4, p.45

  Program ch2ex4.c:
    #include <stdio.h>
    main(void)
    {
      int x;
      x = 37;
      printf("37   = %d(decimal), %o(octal), %x(hexadecimal)\n", x, x, x );
      x = 037;
      printf("037  = %d(decimal), %o(octal), %x(hexadecimal)\n", x, x, x );
      x = 0x37;
      printf("0x37 = %d(decimal), %o(octal), %x(hexadecimal)\n", x, x, x );
      return 0;
    }

  Compilation:
    % cc ch2ex4.c

  Execution:
    % a.out
    37   = 37(decimal), 45(octal), 25(hexadecimal)
    037  = 31(decimal), 37(octal), 1f(hexadecimal)
    0x37 = 55(decimal), 67(octal), 37(hexadecimal)


Exercise 7, p.45

    B = 5; C = 8;
    A = B++ + C++;	/* Now B is 6, C is  9, and A is 5+8  = 13 */
    A = B++ + ++C;	/* Now B is 7, C is 10, and A is 6+10 = 16 */
    A = ++B + C++;	/* Now B is 8, C is 11, and A is 8+10 = 18 */
    A = ++B + ++C;	/* Now B is 9, C is 12, and A is 9+12 = 21 */


Exercise 12, p.45

    int A, B = 5, C = 7;
    double X, Y = 2.8, Z = 16.8;

    /* 1 */ A = Y;			/* Now A is  2  */
    /* 2 */	A = -Y;			/* Now A is -2  */
    /* 3 */	X = A / B + C;		/* Now X is  7, since A/B is -2/5=0 */
    /* 4 */	X = A / B + (double) C;	/* Now X is  7.0000000000000 */
    /* 5 */	X = (double) A / B + C;	/* Now X is  6.6000000000000 */
    /* 6 */	A = Y - Z;		/* Now A is -14 */


Exercise 17, p. 46:

    We assume that our integers have 16 bits.
    First convert from hexadecimal to binary:
    A = 0xC0B3 = 1100 0000 1011 0011 (base 2)
    B = 0x2435 = 0010 0100 0011 0101 (base 2)

    (a)  ~A   =  0011 1111 0100 1100 (base 2) = 0x3F4C   /* affected by bit 
							    length of `int' */
    (b) A | B =  1110 0100 1011 0111 (base 2) = 0xE4B7
    (c) A & B =  0000 0000 0011 0001 (base 2) = 0x0031
    (d) A ^ B =  1110 0100 1000 0110 (base 2) = 0xE486
    (e) A >>2 =  0011 0000 0010 1100 (base 2) = 0x302C   /* affected by sign */
    (f) A <<2 =  0000 0010 1100 1100 (base 2) = 0x02CC   /* affected by bit 
							    length of `int' */
    (g) A<<-2 is undefined, since the right-hand argument is negative.


Extra problem:

    /*  1 */   /* This program contains several errors.	// OK
    /*  2 */    * Can you find them all? */	// Comment ends prematurely
    /*  3 */   #include <stdio.h>		// OK
    /*  4 */   #define Pi=3.141592653589793;	// Use no `=' or `;' in #define
    /*  5 */   main( void )			// OK
    /*  6 */    {				// OK
    /*  7 */      double x, y;			// OK
    /*  8 */      float z;			// OK
    /*  9 */      int i,			// End this statement with `;' 
    /* 10 */      if ( argc > 1 ) {		// `argc' is undeclared.
    /* 11 */         printf('No Arguments!\n');	// Use `"' for strings
    /* 12 */         return(1);			// OK
    /* 13 */      }				// OK
    /* 14 */      x = z = Pi;			// This crashes from bad `Pi'
    /* 15 */      y = x;			// OK
    /* 16 */      printf ( "%f is zero.\n", x-y );	// OK 
    /* 17 */      printf ( "%d is pi.\n", z );		// `z' needs `%f'
    /* 18 */      i = 1/Pi;			// This crashes from bad `Pi'
    /* 19 */      i =* Pi;			// This crashes from bad `Pi'
    /* 20 */      printf ( "%d is one.\n", i );	// OK
    /* 21 */      return(0)			// End statements with `;'
    /* 22 */    }				// OK