Programming Puzzle

1. Declare a double variable named Velocity and a pointer variable named PtrVelocity, and make PtrVelocity refer (point) to variable Velocity.

double Velocity;

double *PtrVelocity;

PtrVelocity = &Velocity;

2. Given this definition: int X_Array[] = {10, 20, 30, 40, 50, 60, 70, 80}; 

What is the meaning or, if applicable, the value of:

a. X_Array 

The address of the first element of the X_Array[] which holds 10 in our case.

b. X_Array + 2 

The address of the third element of the X_Array[] which holds 30 in our case.

c. *X_Array

Since an array is a pointer in itself, *X_Array gives us the value of the first element of the X_Array[] and the value is 10.

d. *X_Array + 2

Since an array is a pointer in itself, *X_Array + 2 adds the value of the first element of the X_Array[] + 2 Hence, the value is 12.

e. *(X_Array + 2)

Since an array is a pointer in itself, *(X_Array + 2) gives us the value of third element of the X_Array[] which is 30 because it is the value in the position that is two bytes ahead of where X_Array starts.

3. A program contains the following declarations and statements.

char u;

char v = 'A';

char* pu;

char* pv = &v;

*pv = v + 1;

u = *pv + 1;

pu = &u;

If the address of u is 5000 and the address of v is 5001, then (next page):

Columbia College - Aurora Campus

CISS 242 Midterm Exam

(a) [2] What value is stored at &v?        _5001 ________

(b) [2] What value is assigned to pv?      _5001_________

(c) [2] What value is represented by *pv?  _65  _________

(d) [2] What value is assigned to u?       _66___________

(e) [2] What value is stored at &u?        _5000_________

(f) [2] What value is assigned to pu?      _5000_________

(g) [2] What value is represented by *pu?  _66___________

4. Write a C-string function named strrev that uses exclusively pointer notation and reverses the elements of a given string. As is the convention of most C-string functions, have your function return the address of the given string.

int strrev (char*);

int strrev (char* str) {

   char* p = str + strlen (str) – 1;

   while (p >= str) {

      cout << *p;

      --p;

   }

   return &str;

}

  

CISS 242 Midterm Exam

5. Write a C-string function named strsum that uses exclusively pointer notation and calculates and returns the sum of the ASCII values of all characters in a given string.

int strsum(char*);

int strsum(char* str)

{

     int num = 0;

     int sum = 0;

     while (str[num] != '\0') {    

         sum += str[num];      

         ++num;

     }

     return sum;

}

6. Write a C-string function named strsumd that uses exclusively pointer notation and calculates and returns the sum of digits embedded within a given string. For example, if the given string were “ab2c38d4e27fg” then the sum would be 26 since 2 + 3 + 8 + 4 + 2 + 7 = 26. (Note: A more difficult function would be to calculate and return the sum of embedded numerical substrings; i.e., for the example given, the sum would be 71 since 2 + 38 + 4 + 27 = 71. Are there any takers for some extra credit?)

int strsumd (char*);

int strsumd (char* str) {

   int sum = 0;

   char* p = str;

   while (*p) {

      if (isdigit (*p)) {

         sum = sum + *p – ‘0’;

      }

   ++p;

   }

   return sum;

} 

7. A challenging “fill-in-the-blanks” take-home problem (next/last page):

MEMORY LOCATIONS - Byte Addresses Shown: Hex Value (Decimal Value)

// Fill in the bits according to the declarations at the bottom.

0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 1 0 1 1 1

0 (0)             1 (1)             2 (2)             3 (3)

1 1 1 1 1 1 1 1   1 1 1 1 1 1 1 1   1 1 1 1 1 1 1 1   1 0 1 0 1 0 1 1

4 (4)             5 (5)             6 (6)             7 (7)

0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0

8 (8)             9 (9)             A (10)            B (11)

0 1 0 0 0 0 1 0   1 0 0 0 0 0 1 0   1 1 1 1 0 1 0 1   1 1 0 0 0 0 1 1

C (12)            D (13)            E (14)            F (15)

1 1 0 0 0 0 0 0   0 0 1 1 1 0 1 1   0 1 1 0 0 0 1 1   1 1 0 1 0 1 1 1

10 (16)           11 (17)           12 (18)           13 (19)

0 0 0 0 1 0 1 0   0 0 1 1 1 1 0 1   0 1 1 1 0 0 0 0   1 0 1 0 0 0 1 1

14 (20)           15 (21)           16 (22)           17 (23)

0 0 1 1 1 1 1 1   0 1 1 1 0 0 1 0   0 1 0 1 0 0 1 0   0 1 0 1 0 0 1 0

18 (24)           19 (25)           1A (26)           1B (27)

0 1 1 0 1 1 1 1   0 1 1 0 1 1 1 0   0 0 0 0 0 0 0 0   0 1 0 0 1 0 0 0

1C (28)           1D (29)           1E (30)           1F (31)

0 1 1 0 0 0 0 1   0 1 1 1 0 0 1 0   0 1 1 1 0 1 0 0   0 0 0 0 0 0 0 0

20 (32)           21 (33)           22 (34)           23 (35)

0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 1 0 0

24 (36)           25 (37)           26 (38)           27 (39)

0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 1 1 0 0

28 (40)           29 (41)           2A (42)           2B (43)

0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 1 1 1 1 1

2C (44)           2D (45)           2E (46)           2F (47)

0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 0 0 0 0 0 0   0 0 1 0 1 1 0 0

30 (48)           31 (49)           32 (50)           33 (51)

_ _ _ _ _ _ _ _   _ _ _ _ _ _ _ _   _ _ _ _ _ _ _ _   _ _ _ _ _ _ _ _

34 (52)           35 (53)           36 (54)           37 (55)

      int Index = 23, Value = -85; // sizeof(int) is 4; 2s-complement for

      long FlightSeconds = 0L;     //                   negative integers.   

      float Velocity = 65.48; // sizeof(float) is 4; this already done.

      double Temp = -27.39; // sizeof(double) is 8; this already done too.

      char Ch = '?', Letter1 = 'r', Letter2 = 'R';

      char FirstName[] = "Ron";   // How many bytes?

      char LastName[]  = "Hart";  // How many bytes?

      int* Valueptr = &Value; // Use 4 bytes for pointer variables.

      float* VelocityPoint = &Velocity;

      char* ByteAddr = LastName;

      char** PointToPoint = &ByteAddr;