1. CECS 130 Mid-term Test Review
CECS 130 Mid-term Exam Review
Summer, 2019
CECS 130 Mid-term Test Review
REACH
Provided by REACH – Resources for Academic Achievement
Presenter:
Nikhil Paonikar

2. To download this presentation…
Go to reach.louisville.edu
Click “Tutoring” at the top left of the page.
Click “Computer Tutoring” under the heading “TUTORING SERVICES FOR ALL STUDENTS.”
Click “CECS Test Reviews” on the right-hand column.
Scroll down to find your class.
Or, just go to tiny.cc/REACH4CECS

3. Variable Types To declare a constant (read only) value:
Data Type
Description
Declaration
Example
Integer
Whole numbers, positive or negative
int x = ;
-3 , 0, 3 , 29
Floating-point number
All numbers, positive or negative, decimals and fractions
float x = ;
, 0.00,
Character
Representations of integer values known as character codes
char x = ‘ ’; m,
To declare a constant (read only) value:
const int x = 20;
const float PI = 3.14;

4. Variable Types TYPE SIZE VALUES bool 1 byte true (1) or false (0) char
‘a’ to‘z’ , ‘A’ to ‘Z’, ‘0’ to ‘9’, space, tab, etc.
int
4 bytes
-2,147,483, to 2,147,483,647
short
2 bytes
-32, to 32,767
long
-2,147,483, to 2,147,483,647
float
+ - (1.2 x 10^-38 to x 10^38)
double
8 bytes
+- (2.3 x 10^ to x 10^308)

5. scanf( ) ; printf( ) int x = 0;
printf(“ What number should I print out? \n ”);
scanf(“%d”, &x); //variables require an “&” sign before them for scanf()!
printf(“\n You chose: %d \n”, x); //but not for printf()!!
Conversion specifiers (%d, %f, etc.) explained in next slide.

6. Conversion Specifiers
Character - %c
Integer %d
Float (decimal)- %f
String - %s
printf Format Tags:
Format: %[flags][width][.precision][length]specifier
Example: %[.precision]specifier
Output:
7 | float fboat = ;
8 | printf( “%.2f, %.3f, %.5f”, fboat, fboat, fboat );
12.12, ,

7. Arithmetic, Logic and Order of Precedence
Operators
Description
( ), !( )
Parentheses: evaluated from innermost to outermost
*, /, %
Multiplication: evaluated from left to right
+ , -
Addition: evaluated from left to right
<, <=, >, >=
Inequalities: evaluated from left to right
==, !=
Equalities: evaluated from left to right
&&, ||
Logical expressions: &&’s first, then ||’s.
*This is a highly abbreviated list. See for the full order of precedence.

8. Predict the printout: User enters ‘2’ and ‘4’:
1 | #include <stdio.h>
2 |
3 | int main()
4 | {
5 | int iNum_1= 0;
6 | int iNum_2= 0;
7 |
8 | printf( “Please enter first number: ” );
9 | scanf( “%d”, &iNum_1 ); // (user enters 2)
10| printf( “\nEnter second number: ” );
11| scanf( “%d”, &iNum_2 ); // (user enters 4)
12| printf( “\n\nThe result is %d. \n”, 24 / (iNum_1 * iNum_2) + 6 / 3);
13|
14| return 0;
15| }
Answer: 5 (shown on next slide)

9. Predict the printout: User enters ‘2’ and ‘4’:
Output:
Please enter first number: 2
Enter second number: 4
The result is 5.
1 | #include <stdio.h>
2 |
3 | int main()
4 | {
5 | int iNum_1= 0;
6 | int iNum_2= 0;
7 |
8 | printf( “Please enter first number: ” );
9 | scanf( “%d”, &iNum_1 );
10| printf( “\nEnter second number: ” );
11| scanf( “%d”, &iNum_2 );
12| printf( “\n\nThe result is %d. \n”, 24 / (iNum_1 * iNum_2) + 6 / 3);
13|
14| return 0;
15| }

10. Can you predict the printout?
1 | #include <stdio.h>
2 |
3 | int main()
4 | {
5 | int x = 9;
6 | int y = 4;
7 | int result1, result2;
8 |
9 | result1 = x/y;
10| result2 = x%y;
11|
12| printf( “\n\nThe result is %d.%d”, result1, 25 * result2);
13| }
Answer: “The result is 2.25” (shown in next slide)

11. Can you predict the printout?
Output:
The result is 2.25
1 | #include <stdio.h>
2 |
3 | int main()
4 | {
5 | int x = 4;
6 | int y = 9;
7 | int result1, result2;
8 |
9 | result1 = y/x;
10| result2 = y%x;
11|
12| printf( “\n\nThe result is %d.%d \n”, result1, 25 * result2);
13| }

12. Conditions and Operators
Description ==
Equal to !=
Not Equal
>
Greater Than
<
Less Than
>=
Greater Than or Equal to
Order of Precedence
Description
&&
AND condition ||
OR condition

13. Comparisons > greater than 5 > 4 is TRUE
< less than 4 < 5 is TRUE
>= greater than or equal 4 >= 4 is TRUE
<= less than or equal 3 <= 4 is TRUE
== equal to 5 == 5 is TRUE
!= not equal to 5 != 4 is TRUE

14. Boolean Operators Do you know the answer to these? A. !( 1 | | 0 )
C. !(( 1 | | 1>0 ) && 0)
= !1 = 0
= !(1||0&&0) = !(1||0) = !1 = 0
= !((1||0)&&0) = !(1&&0) = !0 = 1
(shown in next slide)

15. Boolean Operators Answers: A. B. C. !(1||0) =!(1) =0 !(1||1<0&&0)
=!(1||0&&0)
=!(1||0)
=!(1) =0
!((1||1<0)&&0)
=!((1||0)&&0)
=!(1&&0)
=!(0) =1

16. Example
Using if-statements, write a program that will ask a user
to enter a number 1, 2, or 3, and print out the following:
User Input
Printout 1
Smitty Werbenjagermanjensen was Number 1. 2
Fool me 2 times, can’t put the blame on you. 3
Gimme 3 steps.

17. Example: 1 | #include <stdio.h> 2 | int main() {
4 | printf( “Enter one of the following: %d, %d, or %d\n”,1,2,3 );
5 | scanf( “%d”, &a );
6 | if(a==1||a==2||a==3) {
7 | if(a==1)
8 | printf(“\nSmitty Werbenjagermanjensen was Number %d.\n”, 1);
9 | if(a==2)
10| printf(“\nFool me %d times, can’t put the blame on you.\n”, 2);
11| if(a==3)
12| printf(“\dGimme %d steps.\n”, 3);
10| }
11| else
12| printf(“\nSorry, you entered an invalid value\n” );
16| return 0;

18. Switch-Case Statement
1 | switch ( <var> )
2 | {
3 | case <this-value>:
4 | code to execute if <var> == this-value;
5 | break;
6 | case <that-value>:
7 | code to execute if <var> == that-value;
8 | break;
9 |
10| default:
11| code executed if <var> does not equal any of the values;
12| break;
13| }

19. The while( ) loop
while ( condition ) { Code to execute while the condition is true }
1 | #include <stdio.h>
2 | int main()
3 | {
4 | int x = 0;
5 |
6 | while ( x < 10 )
7 | {
8 | printf( “%d”, x );
9 | x++;
10| printf(“\nFool me %d times, can’t put the blame on you.\n”, 2);
11| }
12| getchar();
13| }

20. The for( ) loop Often used when the # of iterations is already known.
Contains 3 separate expressions:
Variable initialization
Conditional expression
Increment/Decrement
1 | #include <stdio.h>
2 |
3 | int main()
4 | {
5 | int x = 0;
6 | for( x=10; x>=0; x-- ) {
7 | printf( “%d\n”, x );
8 | }
9 |}

21. Break/Continue Statements
Used to exit a loop. Once this statement is executed the program will execute the statement immediately following the end of the loop.
continue;
Used to manipulate program flow in a loop. When executed, any remaining statements in the loop will be skipped and the next iteration of the loop will begin.

22. Function Prototypes & Definitions
Function Prototype Syntax
return-type function_name ( arg_type arg1, ..., arg_type argN )
Function Prototypes tell you the data type returned by the function, the data type of the function’s parameters, how many parameters it takes, and the order of parameters.
Function definitions implement the function prototype
Where are function prototypes located in the program?
Where do you find function definitions?

23. Function Prototypes & Definitions
Where are function prototypes located in the program?
Answer: before the main( ) { } function.
Where do you find function definitions?
Answer: function definitions are self-contained outside of the main( ) { } function, usually written below it.

24. Function Example #include <stdio.h>
int mult (int,int); // function prototype
int main() {
int x;
int y;
int z;
printf( “\n x = "); scanf( "%d", &x );
printf( “\n y = ”); scanf( "%d", &y );
printf( “\n z = ”); scanf( "%d", &z );
printf( “\n x*y = %d\n", mult( x, y ) );
printf( “\n z^2 = %d\n", mult( z, z ) ); }
int mult (int a, int b) { //function definition
return a * b;
Output shown in next slide

25. Function Example Output: x = 2 y = 3 z = 4 x*y = 6 z^2 = 16
#include <stdio.h>
int mult (int,int); // function prototype
int main() {
int x;
int y;
int z;
printf( “ x = " ); scanf( "%d", &x );
printf( “\n y = ”); scanf( "%d", &y );
printf( “\n z = ”); scanf( "%d", &z );
printf( “\n\n x*y = %d\n", mult( x, y ) );
printf( “\n z^2 = %d\n", mult( z, z ) ); }
int mult (int a, int b) //function definition
return a * b;

26. Declaring a 1-D Array
How do you declare a one-dimensional array made up of 10 integers?
Answer: int iArray[10]
Other array declarations:
int iArray[10];
float fAverages[30];
double dResults[3];
short sSalaries [9];
char cName[19]; // 18 characters and 1 null character

27. Declaring a 1-D Array
Why do we initialize variables? Because memory spaces may not be cleared from previous values when arrays are created.
Can initialize an array directly.
E.g.: int iArray[5]={0,1,2,3,4};
Can also initialize an array with a loop such as FOR( )
#include <stdio.h>
main() {
int x;
int iArray[5];
for( x=0; x < 5 ; x++) {
iArray[x] = x; }

28. Example: searching an array
#include <stdio.h>
int main() {
int x, iValue;
int iFound = -1;
int iArray[5];
// initialize the array
for( x=0; x < 5 ; x++) {
iArray[x] = ( x + x ); // array will = { 0, 2, 4, 6, 8 } }
printf(“\n Enter value to search for:”);
scanf(“%d”, &iValue);
// search for number
for(x=0 ; x<5; x++) {
if( iArray[x] == iValue) {
iFound = x;
break;
if(iFound > -1)
printf(“\n I found your search value in element %d \n”, iFound);
else
printf(“\n Sorry, your search value was not found \n”);
return 0;

29. Pointers
Pointer variables, simply called pointers, are designed to hold memory addresses as their values.
Normally, a variable contains a specific value, e.g., an integer, a floating-point value, or a character.
However, a pointer contains the memory address of another variable.

30. Pointer Syntax dataType *pointer_name = &variable_name;
(You can also initialize it “…= NULL;” or “… = 0;”)
It’s important to initialize pointers to prevent fatal runtime errors or accidentally modifying important data.

31. 5 0x3F Pointers int val = 5; //variable declaration
int *val_ptr = &val; //pointer declaration 5
val
0x3F
name:
0x3F
*val_ptr
0x83
name:
Value
Value
location:
location:

32. Pointers
When an ampersand (&) is prefixed to a variable name, it refers to the memory address of this variable.
name:
val
name:
*val_ptr 5
0x3F
Value
Value
location:
&val = 0x3F
location:
&val_ptr = 0x83

33. Ampersand example Output: ????? #include <stdio.h> int main() {
char someChar = 'x';
printf(“%p\n", &someChar);
return 0; }
There’s no way to know what the output will be. It’s the memory address of the variable.
Output: ?????

34. Passing variables with pointers
void exchange(int*, int*);
main() {
int a = 5;
int b = 3;
exchange(&a,&b);
[(3)print a and b] }
//pass by reference
void exchange(int *x, int *y) {
[(1)print *x and *y]
int temp = *i;
int *x = *y;
int *y = temp;
[(2)print *x and *y]
void exchange(int, int);
main() {
int a = 5;
int b = 3;
exchange(a,b);
[(3)print a and b] }
//pass by value
void exchange(int x, int y) {
[(1)print x and y]
int temp = x;
int x = y;
int y = temp;
[(2)print x and y]
Output:
(1) x = 5, y = 3
(2) x = 3, y = 5
(3) a = 3, b = 5
Output:
*x = 5, *y = 3
*x = 3, *y = 5
a = 3, b = 5
Output:
(1) x = 5, y = 3
(2) x = 3, y = 5
(3) a = 5, b = 3

35. Pointers to Arrays
An array variable without a bracket and a subscript represents the starting address of the array.
An array variable is essentially a pointer.
Suppose you declare an array of integer values as follows:
int list[6] = {11, 12, 13, 14, 15, 16};
*(list + 1) is different from *list + 1. The dereference operator (*) has precedence over +.
So, *list + 1 adds 1 to the value of the first element in the array, while *(list + 1) dereferences the element at address list[1] in the array.

36. Pointers to Arrays Output: k = 11 Output: k = 3 main() {
int list[3] = {10, 3, 5};
int k = 0;
k = *list + 1;
printf(“k = %d”, k); }
main() {
int list[3] = {10, 3, 5};
int k = 0;
k = *(list + 1);
printf(“k = %d”, k); }
Order of Operations – the compiler will do addition after the pointer is replaced with it’s referenced value (in the left-hand case above, 10).
That is, unless the addition is within parenthesis. In that case, given *(iArray+n), the pointer references the nth cell in iArray.
Note, one and only one of the variables inside the parenthesis must be an array. Otherwise the compiler throws an error.
Output: k = 11
Output: k = 3

37. Strings Function Description strlen()
Returns numeric string length up to, but not including null character
tolower() and toupper()
Converts a single character to upper or lower case
strcpy()
Copies the contents of one string into another string
strcat()
Appends one string onto the end of another
strcmp()
Compares two strings for equality
strstr()
Searches the first string for the first occurrence of the second string

38. Strings Output: The length of string 1 is 5
#include <stdio.h>
#include <string.h>
int main() {
char *str1 = “REACH”;
char str2[] = “Tutoring”;
printf(“\nThe length of string 1 is %d \n”, strlen(str1));
printf(“\nThe length of string 2 is %d \n”,strlen(str2));
return 0; }
Output:
The length of string 1 is 5
The length of string 2 is 8

39. Strings Output: The name in lowercase is barbara bush
#include <stdio.h>
#include <string.h>
void convertL(char *);
int main() {
char name1[] = “Barbara Bush”;
convertL(name1);
return 0; }
void convertL(char *str)
int x;
for ( x = 0; x <=strlen(str) ; x++)
str[x] = tolower(str[x]);
printf(“\nThe name in lowercase is %s\n”, str);
Output:
The name in lowercase is barbara bush

40. Data File Hierarchy Entity Description Bit Binary digit, 0 or 1
Smallest value in a data file
Byte
Eight bits
Stores a single character
Field
Grouping of bytes
i.e a word, social security number
Record
Grouping of fields
a single row of information, student name, age, ID, GPA
File
Grouping of records
separate fields in a record using spaces, tabs, or commas

41. Data Structures
Arrays require that all elements be of the same data type.
It is often necessary to group information of different data types. An example is a list of materials for a product. The list typically includes a name for each item, a part number, dimensions, weight, and cost.
C and C++ support data structures that can store combinations of character, integer, floating point and enumerated type data.
They are called - structs.

42. Struct Syntax struct oneCar //elements in a list of cars { int year;
char make[10];
char model[10];
char tag[8];
}fleet; //for short we’ll call it a fleet
fleet rentals[5]; // five cars that we rent out to clients
rentals[0].year = 2008;
rentals[0].make = “Honda”;
rentals[0].model = “Element”;
rentals[0].tag = “AS2395”;
[…]
rentals[4].year = 2015;
rentals[4].make = “Tesla”
rentals[4].model = “Model S”;
rentals[4].tag = “2FST4U”;
The blue box is an example of using the same struct for a different fleet of vehicles. In this case, the company wants to manage its company vans separately from the rental cars.
This will apply even more once you start using classes/objects in C++ because they could then add more fields for the vans (mileage, driver, location) vs the rentals (current renter, damages incurred, rental rate, etc.). You don’t need to worry about that for now though.
fleet vans[3]; //company vans
vans[0].year = 2012;
vans[0].make = “Ford”;
vans[0].model = “EconoVan”;
Vans[0].tag = “MV1NUP”;

43. Dynamic memory allocation
These functions are defined in the <stdlib.h> header file.
void *calloc(int num, int size)
Allocates an array of num elements each of whose size in bytes will be size.
void free(void *address)
Releases a block of memory block specified by address.
void *malloc(int num)
Allocates an array of num bytes and leave them uninitialised.
void *realloc(void *address, int newsize)
Re-allocates memory extending it up to newsize.

44. Implementing malloc()

45. Implementing calloc()

46. Resizing and releasing memory using realloc() and free()

47. On the test… Some types of problems you will likely encounter:
True/False questions
Fill-in-the-blanks questions
Finding errors in programs
Common mistakes on this test:
Not appending lines of code with a semicolon (“;”)
Forgetting to use semicolons when writing a FOR loop
Forgetting to add an ampersand (“&”) to your variable parameter in SCANF functions (and mistakenly using the ampersand when using PRINTF)
Mixing up * and & in general.
Off-by-one errors when accessing an array, especially when using a FOR loop.

48. Test Success: The Night Before
Avoid extensive study
Conduct a brief review
Do something relaxing
Get plenty of sleep
Avoid stress-inducing situations
Maintain a positive attitude

49. Test Success: Day of the Test
Get off to a good start.
Arrive on time or early.
Compose yourself.
Do a final revision of your notes.
Maintain a confident attitude.
Eat breakfast if possible.
Compose yourself = deep breaths, calm yourself, get super angry, whatever. Just get in your best test-taking state of mind.
comic by KC Green (

50. Cheers and good luck! Hope for the best, prepare for the worst.
Make sure everyone signed in on the sign in sheet with their name and student ID number!
(If you have not signed in, please do so before you leave!)
This presentation was provided by
REACH – Resources for Academic Achievement

51. To download this presentation…
Go to reach.louisville.edu
bit.ly/31JDZL5
Make sure everyone signed in on the sign in sheet with their name and student ID number!