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**Computer Programming w/ Eng. Applications**

CS 2073 Computer Programming w/ Eng. Applications Ch 2 Simple C Programs Turgay Korkmaz Office: SB Phone: (210) Fax: (210) web:

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Name Addr Content Lecture 3 Lecture++;

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2.1 Program Structure

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**{ } /*-----------------------------------------*/ Comments**

/* Program chapter1_ */ /* This program computes the */ /* distance between two points */ #include <stdio.h> #include <math.h> int main(void) { /* Declare and initialize variables. */ double x1=1, y1=5, x2=4, y2=7, side_1, side_2, distance; /* Compute sides of a right triangle. */ side_1 = x2 - x1; side_2 = y2 - y1; distance=sqrt(side_1*side_1 + side_2*side_2); /* Print distance. */ printf("The distance between the two " "points is %5.2f \n", distance); return 0; /* Exit program. */ } Comments Preprocessor, standard C library every C program must have main function, this one takes no parameters and it returns int value { begin Variable declarations, initial values (if any) Statements must end with ; indentation return 0 } end of function

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General Form 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 a return 0; statement preprocessing directives int main(void) { declarations statements }

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**Another Program /***************************************************/**

/* Program chapter */ /* This program computes the sum of 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); system("pause"); /* keep DOS window on the screen*/ return 0; /* Exit program. */ } /****************************************************/

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**2.2 Constants and Variables**

What is a variable in math? f(x) = x2+x+4 In C, A variable is a memory location that holds a value An identifier or variable name is used to reference a memory location.

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**Memory double x1=1,x2=7,distance; 1 = 00000001 7 = 00000111**

name address Memory - content … 11 12 13 14 15 16 1 = 7 = ? = How many memory cells does your computer have? Say it says 2Gbyte memory? 1K= or 210 = 1024 1M=106 or 220 = 10242 1G= or 230 = 10243 x1 x2 distance

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**Memory Snapshot Name Addr Content x1 1 y1 5 x2 4 y2 7 side_1 ? side_2**

distance

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**Rules for selecting a valid identifier (variable name)**

Must begin with an alphabetic character or underscore (e.g., abcABC_) May contain only letters, digits and underscore (no special characters Case sensitive (AbC, aBc are different) Cannot use C keywords as identifiers (e.g., if, case, while)

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**Are the following valid identifiers?**

initial_time DisTaNce X&Y distance 1x x_1 rate% x_sum switch

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C Numeric Data Types

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**Example Data-Type Limits**

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**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

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Memory name address Memory - content How to represent ‘a’ ? char My_letter=‘a’; int My_number = 97 Always we have 1’s and 0’s in the memory. It depends on how you look at it? For example, is 97 if you look at it as int, or ‘a’ if you look at it as char ‘3’ is not the same as 3 How to represent 2.5? 1 2 3 4 5 6 … ‘a’= 97 = ? = My_letter My_number

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**Program to Print Values as Characters and Integers**

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Constants A constant is a specific value that we use in our programs. For example 3.14, 97, ‘a’, or “hello” In your program, int a = 97; char b =‘a’; double area, r=2.0; double circumference; area = 3.14 * r*r; circumference = 2 * 3.14 * r; ? 2.0 a b area circumference r

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**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

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**2.3 Assignment Statements**

Used to assign a value to a variable General Form: identifier = expression; /* ‘=‘ means assign expression to identifier */ Example 1 double sum = 0; Example 2 int x; x=5; Example 3 char ch; ch = ‘a’; 5 ‘a’ sum x ch

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**Assignment examples (cont’d)**

int x, y, z; x = y = 0; right to left! Z = 1+1; Example 4 y=z; y=5; x y z 2 2 5

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**Assignment examples with different types**

int a, b=5; double c=2.3; … a=c; /* data loss */ c=b; /* no data loss */ ? 5 2.3 2 a b c 5.0 long double, double, float, long integer, integer, short integer, char Data may be lost. Be careful! No data loss

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

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**Exercise: swap First Attempt x=y; y=x; After x=y 5 x y After y=x 5 x y**

3 5 x y Before

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**Exercise: swap Solution temp= x; x=y; y=temp; after temp=x after x=y**

after y = temp 3 temp 5 x y ? temp 3 5 x y Before Will the following solution work, too? temp= y; y=x; x=temp; Write a C program to swap the values of two variables

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Name Addr Content Lecture 4 Lecture++;

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**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

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**Arithmetic Operators (cont’d)**

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 X 5

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**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

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**Implement a program that computes/prints simple arithmetic operations**

Declare a=2, b=5, c=7, d as int Declare x=5.0, y=3.0, z=7.0, w as double d = c%a Print d d = c/a Print d w = z/x Print w d = z/x Print d w = c/a Print w a=a+1 Print a … try other arithmetic operations too..

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**Mixed operations and Precedence of Arithmetic Operators**

int a=4+6/3*2; a=? int b=(4+6)/3*2; b=? a= 4+2*2 = 4+4 = 8 b= 10/3*2 = 3*2= 6 assign = Right to left

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**Extend the previous program to compute/print mixed arithmetic operations**

Declare a=2, b=5, c=7, d as int Declare x=5.0, y=3.0, z=7.0, w as double d = a+c%a Print d d = b*c/a Print d w = y*z/x+b Print w d = z/x/y*a Print d w = c/(a+c)/b Print w a=a+1+b/3 Print a … try other arithmetic operations too..

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**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

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**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; !!! x *= 4+2/3 x = x*4+2/3 wrong x=x*(4+2/3) correct

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**Precedence of Arithmetic Operators (updated)**

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**Writing a C statement for a given MATH formula**

Area of trapezoid area = base*(height1 + height2)/2; How about this

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**Exercise Tension = 2*m1*m2 / (m1 + m2) * g**

wrong Tension = 2*m1*m2 / (m1 + m2) * g 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);

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**Exercise: Arithmetic operations**

? 5 a b c x y 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 = 12 b = 3 c= 5 x = y =

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**Exercise: Arithmetic operations**

Show how C will perform the following statements and what will be the final output? int a = 6, b = -3, c = 2; c= a - b * (a + c * 2) + a / 2 * b; printf("Value of c = %d \n", c);

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**Step-by-step show how C will perform the operations**

output: Value of c = 27

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**Step-by-step show how C will perform the operations**

int a = 8, b = 10, c = 4; c = a % 5 / 2 + -b / (3 – c) * 4 + a / 2 * b; printf("New value of c is %d \n", c);

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**Exercise: reverse a number**

Suppose you are given a number in the range [ ] Write a program to reverse it For example, num is 258 reverse is 852 int d1, d2, d3, num=258, reverse; d1 = num / 100; d2 = num % 100 / 10; d3 = num % 10; reverse = d3*100 + d2*10 + d1; printf(“reverse is %d\n”, reverse); d1 = num / 100; d3 = num % 10; reverse = num – (d1*100+d3) d3*100 + d1;

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Name Addr Content Lecture 5 Lecture++;

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**2.4 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>

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**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

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**Conversion Specifiers for Output Statements**

Frequently Used

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**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

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**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);

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**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

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**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

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**Conversion Specifiers for Input Statements**

Frequently Used

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**Exercise int i; scanf(“%f %i“, &f, &i);**

float f; int i; scanf(“%f %i“, &f, &i); What will be the values stored in f and i after scanf statement if following values are entered 12.5 1 12 1

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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); Otherwise, user will not know what to do! What will happen if you forget & before the variable name?

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**Exercise: How to input two points without re-compiling the program**

printf(“enter x1 y1: “); scanf(“%lf %lf“, &x1, &y1); printf(“enter x2 y2: “); scanf(“%lf %lf“, &x2, &y2);

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Programming exercise Write a program that asks user to enter values for the double variables (a, b, c, d) in the following formula. It then computes the result (res) and prints it with three digits after .

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Exercise Study Section 2.5 and 2.6 from the textbook

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Name Addr Content Lecture 6 Lecture++;

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Library Functions

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**2.7 Math Functions #include <math.h>**

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.

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**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 [-, ].

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**Parameters or Arguments of a function**

A function may contain no argument or contain one or more arguments If more than one argument, list the arguments in the correct order Be careful about the meaning of an argument. For example, sin(x) assumes that x is given in radians, so to compute the sin of 60 degree, you need to first conver 60 degree into radian then call sin function: #define PI theta = 60; theta_rad = theata * PI / 180; b = sin(theta_rad); /* is not the same as sin(theta); */

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**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)); velocity = sqrt(pow(vo,2)+2*a*(x-xo));

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**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 Make sure that a is given in radian; otherwise, first convert it to radian center = (38.19*(pow(r,3)-pow(s,3))*sin(a))/ ((pow(r,2)-pow(s,2))*a); center = (38.19*(r*r*r - s*s*s)*sin(a))/((r*r –s*s)*a);

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**Exercise: Compute Volume**

Write a program to compute the volume of a cylinder of radius r and height h r h

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**Solution: Compute Volume**

Problem Solving Methodology 1. Problem Statement 2. Input/Output Description 3. Hand Example 4. Algorithm Development 5. Testing

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**Solution: Compute Volume (cont’d)**

Problem Statement compute the volume of a cylinder of radius r and height h Input Output Description radius r volume v height h

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**Solution: Compute Volume (cont’d)**

Hand example r=2, h =3, v=37.68 Algorithm Development Read radius Read height Compute Volume Print volume Convert to a program (see next slide)

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**Solution: Compute Volume (coding)**

#include <stdio.h> #define PI int main(void) { /* Declare Variables */ double radius, height, volume; printf("Enter radius: "); scanf("%lf",&radius); printf("Enter height: "); scanf("%lf",&height); /* Compute Volune */ volume = PI*radius*radius*height; /* Print volume */ printf("Volume = %8.3f \n", volume); system("pause"); exit(0); }

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Exercise Write a program to find the radius of a circle given its area. Read area from user. Compute radius and display it. r

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Exercise Write a program that asks user to enter A in degrees, a and b in cm, then computes B=? in degrees C=? in degrees c=? in cm area=? in cm2 A B C a b c For example, given A=36o, a=8 cm, b=5 cm: B=21.55o, C=122.45o, c=11.49 cm

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**Write a program that finds the intersection of two lines and the angle between them**

See handout A1x+B1y+C1=0 A2x+B2y+C2=0

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**2.8 Character Functions #include <ctype.h> putchar(‘a’);**

C= getchar(); 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.

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**Exercise What is the output of the following program**

#include <stdio.h> #include <ctype.h> int main(void) { char ch1='a', ch2; char ch3='X', ch4; char ch5='8'; ch2 = toupper(ch1); printf("%c %c \n",ch1,ch2); ch4 = tolower(ch3); printf("%c %c \n",ch3,ch4); printf("%d\n",isdigit(ch5)); printf("%d\n",islower(ch1)); printf("%d\n",isalpha(ch5)); system("pause"); return(0); }

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Skip Study Section 2.9 from the textbook Skip Section 2.10

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