1. Simple C Programs

2. Program Structure preprocessing directives int main(void) {
declarations
statements }

3. Program Structure
Comments begin with the characters /* and end with the characters */
/* compute distance between two points */
Preprocessor directives give instructions to the compiler
#include <math.h>
Every C program contains one function named main
int main() {
statements }
The body of the main function is enclosed by braces {}

4. Program Structure
The main function contains two types of commands: declarations and statements
Declarations and statements are required to end with a semicolon (;)
Preprocessor directives do not end with a semicolon
To exit the program, use
return 0;

5. Preprocessor Directive
First Program
Comments
/******************************************************************/
/* Program chapter */
/* */
/* This program computes the sum two numbers */
#include <stdio.h>
int main(void) {
/* Declare and initialize variables. */
double number1 = , number2 = 45.7, sum;
/* Calculate sum. */
sum = number1 + number2;
/* Print the sum. */
printf(“The sum is %5.2f \n”, sum);
/* Exit program. */
return 0; }
/*************************************************************/
Preprocessor Directive
No semicolon!
Main
function
Variable Declarations
Statements
Exit the program

6. Constants and Variables
A variable is a memory location that is assigned a name or an identifier
An identifier is used to reference a memory location.
Memory
snapshot

7. Rules for Selecting a valid identifier (variable name)
Must begin with an alphabetic character or underscore
May contain only letters, digits and underscore (no special characters)
Case sensitive
Can not use keywords as identifiers

8. Are the following Valid Identifiers?
initial_time
DisTaNce
X&Y
distance 1x
x_1
rate%
x_sum
switch

9. C Numeric Data Types

10. Example Data-Type Limits

11. C Character Data Type: char
char result =‘Y’;
In memory, everything is stored as binary value, which can be interpreted
as char or integer. Examples of ASCII Codes

12. Symbolic Constants What if you want to use a better estimate of ?
For example, you want instead of 3.14.
You need to replace all by hand
Better solution, define  as a symbolic constant, e.g.
#define PI …
area = PI * r * r;
circumference = 2 * PI * r;
Defined with a preprocessor directive
Compiler replaces each occurrence of the directive identifier with the constant value in all statements that follow the directive

13. a variable of type float
Example
No selicolon!
#define PI
int main() {
int x,y,z;
int a;
int b;
float f1;
double d1;
return 0; }
3 variables of type int
a variable of type float

14. Assignment Statements
Used to assign a value to a variable
General Form:
identifier = expression;
Example 1
double sum = 0; sum
Example 2
int x;
x=5; x
Example 3
char ch;
ch = ‘a’; ch 5 a

15. Assignment Statements
Example 3
int x, y, z;
x=y=0;
z=2; x y z
Example 4
y=z; y 2 2

16. Arithmetic Operators Addition + sum = num1 + num2;
Subtraction - age = 2007 – my_birth_year;
Multiplication * area = side1 * side2;
Division / avg = total / number;
Modulus % lastdigit = num % 10;
Modulus returns remainder of division between two integers
Example 5%2 returns a value of 1
Binary vs. Unary operators
All the above operators are binary (why)
- is an unary operator, e.g., a = -3 * -4

17. Arithmetic Operators
Note that id = exp means assign the result of exp to id, so
x=x+1 means
first perform x+1 and
Assign the result to x
Suppose x is 4, and
We execute x=x+1 4 5 X

18. Integer division vs Real division
Division between two integers results in an integer.
The result is truncated, not rounded
Example:
int A=5/3;  A will have the value of 1
int B=3/6;  B will have the value of 0
To have floating point values:
double A=5.0/3;  A will have the value of 1.666
double B=3.0/6.0;  B will have the value of 0.5

19. Precedence of Arithmetic Operators
Mixed operations:
a=4+6/3*2;  a=?
b=(4+6)/3*2;  b=?
a= 4+2*2 = 4+4 = 8
b= 10/3*2 = 3*2= 6
assign = Right to left

20. Priority of Operators Compute the following Area of trapezoid
2*(3+2)
2*3+2
6-3*2
Area of trapezoid
area = 0.5*height*(base_1+base_2);
base_1
height
base_2

21. Increment and Decrement Operators
Increment Operator ++
post increment x++;
pre increment ++x;
Decrement Operator --
post decrement x--;
pre decrement --x; }
x=x+1; }
x=x-1;
But, the difference is in the following example. Suppose x=10;
A = x++ - 5; means A=x-5; x=x+1; so, A= 5 and x=11
B =++x - 5; means x=x+1; B=x-5; so, B=6 and x=11

22. Abbreviated Assignment Operator
operator example equivalent statement
+= x+=2; x=x+2;
-= x-=2; x=x-2;
*= x*=y; x=x*y;
/= x/=y; x=x/y;
%= x%=y; x=x%y;

23. Precedence of Arithmetic Operators (updated)

24. Exercise Write a C statement to compute the following
f = (x*x*x-2*x*x+x-6.3)/(x*x+0.05*x+3.14);

25. Exercise
Write a set of statements that swaps the contents of variables x and y 3 5 5 3 x y x y
Before
After

26. Exercise First Attempt x=y; y=x; 3 5 5 5 5 5 x y x y x y Before
After x=y
After y=x

27. Exercise Solution temp= x; x=y; y=temp; 3 5 ? 3 5 3 5 5 3 5 3 3 x y
Before
after temp=x
after x=y
after y = temp

28. Standard Input and Output

29. Standard Input and Output
Output: printf
Input: scanf
Remember the program computing the distance between two points!
/* Declare and initialize variables. */
double x1=1, y1=5, x2=4, y2=7,
side_1, side_2, distance;
How can we compute distance for different points?
It would be better to get new points from user, right? For this we will use scanf
To use these functions, we need to use
#include <stdio.h>

30. Standard Output printf Function prints information to the screen
requires two arguments
control string
Contains text, conversion specifiers or both
Identifier to be printed
Example
double angle = 45.5;
printf(“Angle = %.2f degrees \n”, angle);
Output:
Angle = degrees
Conversion Specifier
Control String
Identifier

31. Conversion Specifiers for Output Statements
Frequently Used

32. Standard Output Output of -145 Output of 157.8926 Specifier
Value Printed %i
-145
%4d
%3i
%6i
__-145
%-6i
-145__
%8i
____-145
%-8i
-145____
Specifier
Value Printed %f
%6.2f
157.89
%7.3f
%7.4f
%7.5f %e
e+02
%.3E
1.579E+02

33. Exercise int sum = 65; double average = 12.368; char ch = ‘b’;
Show the output line (or lines) generated by the following statements.
printf("Sum = %5i; Average = %7.1f \n", sum, average);
printf("Sum = %4i \n Average = %8.4f \n", sum, average);
printf("Sum and Average \n\n %d %.1f \n", sum, average);
printf("Character is %c; Sum is %c \n", ch, sum);
printf("Character is %i; Sum is %i \n", ch, sum);

34. Exercise (cont’d) Solution Sum = 65; Average = 12.4 Sum = 65
Sum and Average
Character is b; Sum is A
Character is 98; Sum is 65

35. Standard Input scanf Function inputs values from the keyboard
required arguments
control string
memory locations that correspond to the specifiers in the control string
Example:
double distance;
char unit_length;
scanf("%lf %c", &distance, &unit_length);
It is very important to use a specifier that is appropriate for the data type of the variable

36. Conversion Specifiers for Input Statements
Frequently Used

37. Exercise float f1; int i1; scanf(“%f %d”,&f1,&i1);
f1 is float and conversion specifier for float is f
i1 is integer and conversion specifier for integer is %f
float f1;
int i1;
scanf(“%f %d”,&f1,&i1);
What will be the values stored in f and i after scanf statement if following values are entered
12.5 1
12 45 12 1

38. Program #include <stdio.h> int main() { float f1; int i1;
scanf("%f %i",&f1,&i1);
printf("f=%f i=%i\n",f1,i1);
system(“pause”);
return(0); }

39. Good practice
You don’t need to have a printf before scanf, but it is good to let user know what to enter:
Printf(“Enter x y : ”);
Scanf(“%d %d”, &x, &y);
User may not know what to do of you have a scanf statement only
What will happen if you forget the & before the variable name?

40. Exercise: Read two points from user
printf(“enter x1 y1: “);
scanf(“%lf %lf“, &x1, &y1);
printf(“enter x2 y2: “);
scanf(“%lf %lf“, &x2, &y2);

41. Library Functions

42. Math Functions fabs(x) Absolute value of x.
sqrt(x) Square root of x, where x>=0.
pow(x,y) Exponentiation, xy. Errors occur if
x=0 and y<=0, or if x<0 and y is not an integer.
ceil(x) Rounds x to the nearest integer toward  (infinity).
Example, ceil(2.01) is equal to 3.
floor(x) Rounds x to the nearest integer toward - (negative infinity). Example, floor(2.01) is equal to 2.
exp(x) Computes the value of ex.
log(x) Returns ln x, the natural logarithm of x to the base e. Errors occur if x<=0.
log10(x) Returns log10x, logarithm of x to the base 10.
Errors occur if x<=0.

43. Trigonometric Functions
sin(x) Computes the sine of x, where x is in radians.
cos(x) Computes the cosine of x, where x is in radians
tan(x) Computes the tangent of x, where x is in radians.
asin(x) Computes the arcsine or inverse sine of x,
where x must be in the range [-1, 1].
Returns an angle in radians in the range [-/2,/2].
acos(x) Computes the arccosine or inverse cosine of x,
Returns an angle in radians in the range [0, ].
atan(x) Computes the arctangent or inverse tangent of x. The Returns an angle in radians in the range [-/2,/2].
atan2(y,x) Computes the arctangent or inverse tangent of the value y/x. Returns an angle in radians in the range [-, ].

44. Exercise velocity = sqrt(vo*vo+2*a*(x-xo));
Write an expression to compute velocity using the following equation
Assume that the variables are declared
velocity = sqrt(vo*vo+2*a*(x-xo));

45. Exercise center = (38.19*(pow(r,3)-pow(s,3))*sin(a))/
Write an expression to compute velocity using the following equation
Assume that the variables are declared
center = (38.19*(pow(r,3)-pow(s,3))*sin(a))/
((pow(r,2)-pow(s,2))*a);

46. Memory Requirement of Types
Integer Types
short: 2 bytes
int: 4 bytes
long: 4 bytes
Floating point Types = *102
float: 4 bytes
double: 8 bytes
long double: 12 bytes
See chapter2e2.c 16

47. Exercise: Arithmetic operations? 5
Show the memory snapshot after the following operations by hand
int a, b, c=5;
double x, y;
a = c * 2.5;
b = a % c * 2 - 1;
x = (5 + c) * 2.5;
y = x – (-3 * a) / 2;
Write a C program and print out the values of a, b, c, x, y and compare them with the ones that you determined by hand. a b c x y

48. Solution #include <stdio.h> int main() { int a, b, c=5;
double x, y;
a = c * 2.5;
b = a % c * 2 - 1;
x = (5 + c) * 2.5;
y = x - (-3 * a) /2;
printf("a = %d b = %d c= %d x = %5.4f y = %5.4f \n", a, b, c, x, y);
system("pause");
return 0; }
a = 12 b = 3 c= 5 x = y =

49. Character Functions
toupper(ch) If ch is a lowercase letter, this function returns the corresponding uppercase letter; otherwise, it returns ch
isdigit(ch) Returns a nonzero value if ch is a decimal digit; otherwise, it returns a zero.
islower(ch) Returns a nonzero value if ch is a lowercase letter; otherwise, it returns a zero.
isupper(ch) Returns a nonzero value if ch is an uppercase letter; otherwise, it returns a zero.
isalpha(ch) Returns a nonzero value if ch is an uppercase letter or a lowercase letter; otherwise, it returns a zero.
isalnum(ch) Returns a nonzero value if ch is an alphabetic character or a numeric digit; otherwise, it returns a zero.

50. Exercise Given a 3 digit number in the range [100.999]
Write a program to reverse it
For example,
num is 258, reverse is 852
num is 123, reverse is 321
num is 111, reverse is 111
How to do it
First approach: Swap first and last digit
Second approach: Split number into digits and combine them in reverse order

51. Exercise continued Suppose you have number 258
How can you compute the last digit?
Find remainder when you divide 258 by 10
d3 = 258 % > 8
How can you compute the first digit?
Divide 258 by 100 and keep integer part
d1 = 258 / > 2
How can you compute the middle digit?
Get the last 2 digits by finding remainder after division by 100
Divide last 2 digits by 10 and keep the integer part
d2 = 258 % 100 / 10 -> 58/10 -> 5
Construct a reverse number with the digits
number2 = d3*100+d2*10+d1 -> 8*100+5*10+2 -> 852