Arithmetic and logical operators
No. 34
Calculating a circle's area
Q: 
The area
$a$ of a given circle having radius
$r$ is being obtained by
$a=\pi \times {r}^{2}$. Complete the following code to calculate
the result and write it to standard output using
TipYou may want to read the overview section on statements in [Kurniawan]. 
A: 

No. 35
Dividing values
Q: 
Consider the following statement:
The output is 
A: 
The divide operator acts on two literals 8 an 9 of type
According to 
No. 36
Strange things with operator ++
Q: 
TipYou may want to read the overview section on statements in [Kurniawan]. 

A: 
Conclusion: 
No. 37
Adding values
Q: 
Consider the following code:
This yields: 2147483648 2147483648 Explain this result. Please refrain from answering “Thats what bankers do!” 
A: 
The value 2147483647 is actually the largest possible
On binary level adding an 01111111_11111111_11111111_11111111 2147483647 + 00000000_00000000_00000000_00000001 + 1   10000000_00000000_00000000_00000000 2147483648 With respect to twocomplement representation of signed
On contrary the plus operator in the expression 01111111_11111111_11111111_11111111 2147483647 + 00000000_00000000_00000000_00000000_00000000_00000000_00000000_00000001 + 1   00000000_00000000_00000000_00000000_10000000_00000000_00000000_00000000 2147483648 Due to a 
No. 38
Representational float and double miracles
Q: 
Consider and execute the following code snippet:
Which outcome do you expect? Explain the execution's result and propose a solution. Tip


A: 
The expression Adding Comparing
The last line represents the boolean expression $\leftxy\right<{\mathrm{10}}^{\mathrm{14}}$. So two values will be regarded as equal if their mutual distance is less than 0.00000000000001. 
try {
int sum = Math.addExact(2147480000, 2147480000);
System.out.println("sum = " + sum);
} catch (ArithmeticException ex) {
System.err.println("Problem: " + ex.getMessage());
}
Problem: integer overflow

Value: Infinity 
Watch out: “Silent” error!
No. 39
Expressions involving infinity
Q: 
Figure 114, “Dividing by zero ” raises some interesting questions:

A: 

Get a division's remainder:
int nuggets = 11,
diggers = 3;
System.out.println("Nuggets per digger:" + nuggets / diggers);
System.out.println("Remaining nuggets:" + nuggets % diggers); 
Nuggets per digger:3 Remaining nuggets:2 
Left  Right  Out  Examples 

boolean 
boolean 
boolean 
, &, &&, , ^ 
int 
int 
int 
+, , *, /, % (read here!) 
int 
long 
long 

byte 
byte 
int 

double 
float 
double 

int 
float 
float 

char 
byte 
int 
The careful reader may
stumble upon the absence of e.g. a binary «+» operator turning two byte
values a
and b
into an expression a + b
of type byte
rather than int
:
byte a = 1, b = 2;
byte sum = a + b; // Error: Incompatible types. ❶
// Required: byte
// Found: int
This is due to a Java Virtual Machine design decision leading to a limited set of computational types:

No. 40
int
to short
assignment
Q: 
Consider the following code segment:
On contrary the following statement compiles flawlessly: Figure 120. Constant expression assignment
Explain this strange behaviour. 

A: 
Considering The expression 
No. 41
int
to short
assignment using final
Q: 
We reconsider a variant of Figure 119, “ final short a = 4;
short sum = a + 7; This time the code compiles flawlessly. Explain the
underlying reason being related to the 
A: 
A final variable's value cannot change. Thus the
expression a + 7 is fully equivalent to 4 + 7 and can thus be
evaluated at compile time. Like in the
preceding Figure 120, “Constant expression assignment”
code sample the resulting value of 
No. 42
Calculating a circle's area avoiding accidental redefinition
Q: 
In Exercise Calculating a circle's area you calculated a given circle's area:
Though there is nothing wrong with this approach it is
error prone: A careless programmer might accidentally redefine
the value of
Modify the above code to avoid this type of error. Tip


A: 
The solution is straightforward. We just add a final ❶ double pi = 3.141592653589793; ... pi = 4; ❷ ...
As a rule of thumb: Whenever you intend a variable not to
change after an initial assignment use In addition refactoring our variable

No. 43
Turning weeks into seconds
Q: 
Consider:
Create an expression based on the variables After you have completed this task read about Horner's rule and try to improve your solution with respect to execution time. 
A: 
A straightforward solution reads:
This requires four additions and 10 multiplications. Horner's rule allows for:
This still leaves us with four additions but just 4 instead of 10 multiplications. Within the given context the difference in execution time can be neglected. But in a larger application the calculation in question might have to be repeated millions of times giving rise to a substantial gain with respect to execution time. 
No. 44
Turning seconds into weeks
Q: 
This exercise is about reversing Turning weeks into seconds namely decomposing a given value of elapsed seconds into weeks, days, hours, minutes and seconds:
TipAs an example consider decomposing 192 seconds into minutes and seconds:
Thus 192 seconds equal 3 minutes and 12 seconds 
A: 
We extend the given minute calculation example to all 5 desired values:

No. 45
Using predefined Java™ standard library constants
Q: 
In the previous exercise we coded:
Do we actually have to provide the value of pi (3.141592653589793) ourself? TipConsider the standard Math library. 
A: 
The solution is straightforward using Math.PI:

No. 46
Converting temperature values
Q: 
Write an application converting temperature values being represented as degree centigrade to kelvin and Fahrenheit:

A: 

No. 47
Time unit conversion
Q: 
This is a two part exercise:

A: 

No. 48
Interest calculation
Q: 
We want to calculate the compounded interest starting from an initial capital, a given annual interest rate and a duration of three years. Consider the following code fragment:
The expected output is: Initial capital: 223.12€ Annual interest rate: 1.5% Interest period: 3 years Capital after three years:233.31175902999993 Extend the given code accordingly. TipIn case you are unsure read about calculating compounded interest. In case you already have prior knowledge about loop statements in Java™ you may want to allow for a variable number of years: // Interest parameters
final double initialCapital = 223.12;
final double interestRate = 1.5;
final int interestPeriodInYears = 3;
...
// Printing result
System.out.println("Capital after three years:" + finalCapital); 
A: 
We obtain the three year's compounded interest by:
$$\mathrm{finalCapital}=\mathrm{initialCapital}{\left(1+\frac{\mathrm{interestRate}}{\mathrm{100}}\right)}^{3}$$
Since we have not yet introduced loops the multiplication has to be repeated three times ❶:
We might as well use a single arithmetic expression achieving the same result:
Using loop statements our solution becomes more flexible with respect to the interest period in question:
$$\mathrm{finalCapital}=\mathrm{initialCapital}{\left(1+\frac{\mathrm{interestRate}}{\mathrm{100}}\right)}^{\mathrm{interestPeriodInYears}}$$
In the section called “Interest calculations” we will present an even more elaborate solution based on classes. 
No. 49
Summing short and char
Q: 
Consider the following snippet:
Execution results in TipTry to assign the expression to another variable and experiment using type candidates. Try to make an educated guess explaining the result. 
A: 
Both
According to the compiler's error message the expression
The two input types have different ranges:
A promotion to 
Boolean “and” of two operands:
boolean examSuccess = true,
registered = false;
boolean examPassed = examSuccess & registered;
System.out.println("Exam passed:" + examPassed); 
Exam passed:false 
No. 50
Operator & vs. &&
Q: 
Execute the following snippet:
Note the resulting output. Then replace the operator TipRead 15.22.2
and 15.23 to understand the difference between 
A: 
Execution yields: Exam success:false, points = 5 On modifying our code by Exam success:false, points = 4 The operators The The Likewise the 