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