Introduction to Computer Programming, Spring Term 2018 Practice Assignment 1 Discussion:

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1 German University in Cairo Media Engineering and Technology Prof. Dr. Slim Abdennadher Dr. Rimon Elias Dr. Hisham Othman Introduction to Computer Programming, Spring Term 2018 Practice Assignment 1 Discussion: Exercise 1-1 Java Expressions To be discussed in the Lab a) ( 1) Syntactically correct, value 1. b) 1 Syntactically incorrect. is an operator in Java that only works on variables. Correct uses of this operator (and the related ++ operator) are: ++i, i, i++, and i. c) 2 4( 6) Syntactically incorrect. There is an operator missing between 4 and (-6). d) (12) (4) 6 Syntactically correct, value 42. e) 29%5%3 4 2 Syntactically correct, value 2. f) 29%5%3 4 2 Syntactically correct, value -7. g) 7%3 = 1 Syntactically Incorrect. The assignment operator operator ==. = requires a variable on the left side. Danger of confusion with the equality h) 1 / 0 Syntactically correct, run-time error. 1

2 i) 1 / 0.0 Syntactically correct, value infinity. j) ((10 2) 4%)8) Syntactically incorrect, ) expected. k) c h a r c = a ; i n t x = c ; System. o u t. p r i n t l n ( x ) ; x ++; System. o u t. p r i n t l n ( x ) ; c h a r v = ( c h a r ) x ; System. o u t. p r i n t l n ( c ) ; System. o u t. p r i n t l n ( v ) ; c ++; System. o u t. p r i n t l n ( c ) ; Syntactically correct. l) f l o a t x = 2. 5 ; f l o a t y = 3. 4 ; f l o a t z = x + y ; System. o u t. p r i n t ( " The v a l u e i s " + z ) ; Syntactically incorrect. By default in Java, a floating point number is represented as double with 64-bits. Thus, we can not put a double in float without writing the f suffix. m) i n t x = ; long y = ; long z = x + y ; System. o u t. p r i n t l n ( " The v a l u e i s " + z ) ; y = ; z = x + y ; System. o u t. p r i n t ( " The v a l u e i s " + z ) ; Syntactically incorrect. By default in Java, a literal for integers is an int with 32-bits. If we want to store a number like that exceeds the range of the int, we need to write the l suffix.. n) b y t e x = 1 2 ; b y t e y = 8 ; b y t e z = x + y ; System. o u t. p r i n t ( " The v a l u e i s " + z ) ; Syntactically incorrect. On the hardware level the + operation is done on int values. Thus, we need to typecast it into byte as well. 2

3 o) S t r i n g x = "CSEN " ; i n t y = 2 02; S t r i n g z = x + y + "! " ; System. o u t. p r i n t l n ( z ) ; S t r i n g u = y + x + "! " ; System. o u t. p r i n t l n ( u ) ; S t r i n g v = y ; System. o u t. p r i n t l n ( v ) ; S t r i n g m = " " ; i n t o = m; System. o u t. p r i n t l n ( o ) ; Syntactically incorrect. int can not be converted into String, we need to concatenate it to a String. Also, String can not be converted into int directly, we need to parse it. Exercise 1-2 Errors and Conventions To be discussed in the lab In the following lab exercises, follow the instructions carefully and observe the outcome. In each case you should be able to explain why the output is as it is. You are encouraged to check out the useful links on MET Website For each of the.java classes in the Lab 1 Exercises.zip zipped folder, compile the files on JCreator. You are asked to define in each file the lines which have the following faults: a) Syntax Errors b) Runtime Errors c) Logic Errors Furthermore, specifically define the errors. For example, the following line of code: String Word = "Hello World!"; system.out.print(word) has two Syntax Errors and 1 bad identifier naming. The correct solution is: String word = "Hello World!"; System.out.println(word); Exercise 1-3 Number Precision, Exactness Given the snippet of code below, what will be the values of d3 and d4 that will be printed? double d1 = ; double d2 = ; double d3 = d1 d2 ; System. o u t. p r i n t l n ( d3 ) ; double d4 = 0. 1 ; f l o a t f = ( f l o a t ) d4 ; d4 = 1 f ; System. o u t. p r i n t l n ( d4 ) ; 3

4 d3 will equal , and d4 will equal Exercise 1-4 Time To be discussed in Tutorial Write an algorithm that reads the amount of time in seconds and then displays the equivalent hours, minutes and remaining seconds. One hour corresponds to 60 minutes. One minute corresponds to 60 seconds. import j a v a. u t i l. Scanner ; p u b l i c c l a s s Time p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) i n t seconds, minutes, h o u r s ; System. o u t. p r i n t l n ( " E n t e r number of s e c o n d s " ) ; s e c o n d s = sc. n e x t I n t ( ) ; h o u r s = s e c o n d s / 3600; s e c o n d s = s e c o n d s % 3600; m i n u t e s = s e c o n d s / 6 0 ; s e c o n d s = s e c o n d s % 6 0 ; System. o u t. p r i n t l n ( " Hours : " + h o u r s ) ; System. o u t. p r i n t l n ( " Minutes : " + m i n u t e s ) ; System. o u t. p r i n t l n ( " Seconds : " + s e c o n d s ) ; Exercise 1-5 Supermarket Change To be discussed in Tutorial Write a Java program CountChange to count change. Given the number of quarter, dimes, nickles, and pennies the program should output the total as a single value in dollars and pennies. Hint: One dollar corresponds to 100 pennies. One quarter corresponds to 25 pennies. One dime corresponds to 10 pennies. One nickle corresponds to 5 pennies. For example if we have: 3 quarters, 2 dimes, 1 nickle and 6 pennies, then the total is 1.06 dollars. Initializing variables by value: 4

5 p u b l i c c l a s s CountChange p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) i n t q u a r t e r s, dimes, n i c k l e s, p e n n i e s ; q u a r t e r s = 3 ; dimes = 2 ; n i c k l e s = 1 ; p e n n i e s = 6 ; double t o t a l = q u a r t e r s 25 + dimes 10 + n i c k l e s 5 + p e n n i e s ; System. o u t. p r i n t l n ( " T o t a l i s : $ " + t o t a l / ) ; Initializing variables by input: import j a v a. u t i l. Scanner ; p u b l i c c l a s s CountChange p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) i n t q u a r t e r s, dimes, n i c k l e s, p e n n i e s ; System. o u t. p r i n t l n ( " E n t e r number of q u a r t e r s " ) ; q u a r t e r s = sc. n e x t I n t ( ) ; System. o u t. p r i n t l n ( " E n t e r number of dimes " ) ; dimes = sc. n e x t I n t ( ) ; System. o u t. p r i n t l n ( " E n t e r number of n i c k l e s " ) ; n i c k l e s = sc. n e x t I n t ( ) ; System. o u t. p r i n t l n ( " E n t e r number of p e n n i e s " ) ; p e n n i e s = sc. n e x t I n t ( ) ; double t o t a l = q u a r t e r s 25 + dimes 10 + n i c k l e s 5 + p e n n i e s ; System. o u t. p r i n t l n ( " T o t a l i s : $ " + t o t a l / ) ; Exercise 1-6 Cook in a Hurry To be discussed in Lab You want to cook some pasta, but you are short of time. To be sure you can finish cooking before your next appointment, you need to know how long it takes for the water to boil. At its highest setting, your stove needs two minutes per liter (1l = 1 1,000 m3 ) to reach the boiling point. You use a cylindrical pot. Write a program that, given the diameter of the pot and the height of the water in it, calculates the the time needed for the water to boil. Hint. The Volume of a cylinder is π r 2 h. p u b l i c c l a s s CookSomePasta p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) f i n a l double PI = ; double d i a m e t e r = 7. 0 ; double h e i g h t = 6. 0 ; double r a d i u s = d i a m e t e r / 2 ; / / c a l c u l a t e t h e volume double volume = PI r a d i u s r a d i u s h e i g h t ; 5

6 / / Assume t h a t d i a m e t e r & h e i g h t are g i v e n i n c e n t i m e t e r s double l i t e r s = volume / 1000; / / The s t o v e needs two m i n u t e s per l i t e r double time = l i t e r s 2 ; System. o u t. p r i n t ( " Stove needs " + time + " m i n u t e s. " ) ; import j a v a. u t i l. ; p u b l i c c l a s s CookSomePasta p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) f i n a l double PI = ; System. o u t. p r i n t ( " E n t e r d i a m e t e r : " ) ; double d i a m e t e r = sc. nextdouble ( ) ; System. o u t. p r i n t ( " E n t e r h e i g h t : " ) ; double h e i g h t = sc. nextdouble ( ) ; double r a d i u s = d i a m e t e r / 2 ; / / c a l c u l a t e t h e volume double volume = PI r a d i u s r a d i u s h e i g h t ; / / Assume t h a t d i a m e t e r & h e i g h t are g i v e n i n c e n t i m e t e r s double l i t e r s = volume / 1000; / / The s t o v e needs two m i n u t e s per l i t e r double time = l i t e r s 2 ; System. o u t. p r i n t ( " Stove needs " + time + " m i n u t e s. " ) ; Exercise 1-7 Electrical Resistance The equivalent resistance of resistors connected in series is calculated by adding the resistances of the individual resistors. The formula for resistors connected in parallel is a little more complex. Given two resistors with resistances R1 and R2 connected in parallel the equivalent resistance is given by the inverse of the sum of the inverses; e.g. 1 Req = 1 R1 + 1 R2. Write a program that given 3 resistances outputs the equivalent resistance when they are connected in series and when they are connected in parallel. p u b l i c c l a s s R e s i s t a n c e p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) throws IOException f l o a t r1 = 8, r2 = 8, r3 = 4 ; f l o a t r e s S = r1 + r2 + r3 ; f l o a t r e s P I n v = 1 / r1 + 1 / r2 + 1 / r3 ; f l o a t r e s P = 1 / r e s P I n v ; 6

7 System. o u t. p r i n t l n ( " S e r i e s r e s i s t a n c e : " + r e s S ) ; System. o u t. p r i n t l n ( " P a r a l l e l r e s i s t a n c e : " + r e s P ) ; Exercise 1-8 Temperature Conversion Assume that you have a Celsius scale temperature of 100 degrees and you wish to convert it into degrees on the Fahrenheit scale and the Kelvin scale. This conversion is done using the following formulas. T f = 9 5 T c + 32, and T k = T c Where T f is temperature in degrees Fahrenheit, T c is temperature in degrees Celsius, and T k is temperature in degrees Kelvin. Write a Java program that takes as an input a degree in Celsius scale and converts it into both Fahrenheit and Kelvin scales. import j a v a. u t i l. ; p u b l i c c l a s s T e m p e r a t u r e C a l c u l a t o r p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) double c e l s i u s, f a h r e n h e i t, k e l v i n ; System. o u t. p r i n t ( " P l e a s e i n s e r t t h e t e m p e r a t u r e i n d e g r e e s C e l s i u s " ) ; c e l s i u s = sc. nextdouble ( ) ; f a h r e n h e i t = ( 9. 0 / 5 c e l s i u s ) ; k e l v i n = c e l s i u s ; System. o u t. p r i n t l n ( " Temperature i n d e g r e e s F a h r e n h e i t i s " + f a h r e n h e i t ) ; System. o u t. p r i n t l n ( " Temperature i n d e g r e e s Kelvin i s " + k e l v i n ) ; Exercise 1-9 Octal Conversion Assume that you have an octal number (base 8) and you would like to convert it into its equivalent decimal number (base 10) and binary number (base 2). For example: (27) 8 is equal to (23) 10 in decimal and (10111) 2 in binary. Write a sequential Java program to convert a two digit octal number into its equivalent binary and decimal numbers. import j a v a. u t i l. ; p u b l i c c l a s s O c t a l p u b l i c s t a t i c void main ( S t r i n g [ ] a r g s ) i n t oct0, o c t 1 ; i n t q0, q1, q2, r0, r1, r2 ; System. o u t. p r i n t ( " P l e a s e e n t e r t h e o c t a l number : " ) ; i n t o c t a l = sc. n e x t I n t ( ) ; 7

8 / / e x t r a c t i n g t h e d i g i t s o c t 0 = o c t a l %10; o c t 1 = o c t a l / 1 0 ; / / C a l c u l a t e t h e d e c i m a l Value double dec = o c t o c t 1 8 ; i n t r e s u l t = ( i n t ) dec ; System. o u t. p r i n t l n ( o c t a l + " i n o c t a l = " + r e s u l t + " i n d e c i m a l. " ) ; / / c o n v e r t i n g t h e f i r s t d i g i t base 8 i n t o 3 b i t s q0 = o c t 1 / 2 ; r0 = o c t 1 %2; q1 = q0 / 2 ; r1 = q0%2; q2 = q1 / 2 ; r2 = q1%2; System. o u t. p r i n t ( o c t a l + " i n o c t a l = "+ r2 +" "+ r1 +" "+ r0 +" " ) ; / / c o n v e r t i n g t h e second d i g i t base 8 i n t o 3 b i t s q0 = o c t 0 / 2 ; r0 = o c t 0 %2; q1 = q0 / 2 ; r1 = q0%2; q2 = q1 / 2 ; r2 = q1%2; System. o u t. p r i n t l n ( r2 +" "+ r1 +" "+ r0 +" i n b i n a r y. " ) ; 8