Unary operators

Figure 122. Increment operator ++

Increment variable by 1:	Shorthand version:
int a = 4; a = a + 1; System.out.println("Value:" + a);	int a = 4; a++; System.out.println("Value:" + a);
Value:5

Figure 123. Different range behaviour!

Increment variable by 1:	Shorthand version:
byte value = 127; // Max possible value value = value + 1; // Error: Required type: byte // Provided:int System.out.println(value);	byte value = 127; // Max possible value value++; // o.K., will cycle through range System.out.println(value);
Does not compile	Value:-128

Increment variable by 1:

Shorthand version:

byte value = 127; // Max possible value

value = value + 1; // Error: Required type: byte
                   // Provided:int

System.out.println(value);

byte value = 127; // Max possible value

value++; // o.K., will cycle through range

System.out.println(value);

Does not compile

Value:-128

Figure 124. Cast required

Increment variable by 1:	Shorthand version:
byte value = 127; // Max possible value value = (byte)(value + 1); // cast required, // possible overflow System.out.println(value);	byte value = 127; // Max possible value value++; // cycle through range, // no cast required. System.out.println(value);
Value:-128

Increment variable by 1:

Shorthand version:

byte value = 127; // Max possible value

value = (byte)(value + 1); // cast required,
                           // possible overflow

System.out.println(value);

byte value = 127; // Max possible value

value++; // cycle through range,
         // no cast required. 

System.out.println(value);

Value:-128

Figure 125. Prefix and postfix notation

pre-increment		post-increment
int a = 4; int b = ++a; System.out.println(b);		int a = 4; int b = a++; System.out.println(b);
Output:	5	Output:	4

exercise No. 52

Three ways expressing the same

Q:	Write the statement `a = a - 1;` in two alternate ways. Tip Use different operators.
A:	`a -= 1;` `a--;`

Figure 126. Operator examples

Shorthand Operation Explicit

Shorthand	Operation	Explicit
//Integer operations i--; i += k; i %= 3; // boolean - nothing appropriate -	`Decrement by one Raise by k's value assign modulus of 3 Switching true <--> false`	//Integer operations i = i - 1; i = i + k; i = i % 3; // boolean b = !b;

//Integer operations
i--;    
i += k;
i %= 3;

// boolean
  - nothing appropriate -

    Decrement by one
    Raise by k's value
   assign modulus of 3



Switching true <--> false

//Integer operations
i = i - 1;
i = i + k;
i = i % 3;

// boolean
b = !b;

exercise No. 53

Guessing results

Consider the following code segment:

int a = 3;
a++;          // Incrementing a by 1 --> a==4

int b = a;    // TODO

b--;          // TODO
--b;          // TODO

int c = b;    // TODO

b = ++a;      // TODO
int e = a++;  // TODO

a *= b;       // TODO

System.out.println("a=" + a);
System.out.println("b=" + b);
System.out.println("c=" + c);
System.out.println("e=" + e);

Guess the expected result without executing the above code. Replace the TODO sections by explanations among with expected variable values.

Then copy/paste the above code into a main() method body and control your findings (possibly wailing about your success rate).

Tip

Both x++ (postfix notation) and ++x (infix notation) are expressions themselves which might even be rewritten as (x++) and (++x) for the sake of clarity. The difference is not about operator precedence rules but simply about the values of these expressions when being assigned to other variables.

As inferred by the hint the biggest problem is about understanding postfix and infix notation of the operators ++ and --. A corresponding expression evaluates to:

a = x++ yields a =x: The value of x before being incremented.
a = ++x yields a = x + 1: The value of x after being incremented.

The remaining task is just obeying the “due diligence” rule set:

int a = 3;
a++;          //Incrementing a by 1 --> a==4

int b = a;    // Assigning value of a --> b==4

b--;          // Decrementing b by 1 --> b==3
--b;          // Decrementing b by 1 --> b==2

int c = b;    // c == 2;

b = ++a;      // Incrementing a by 1 --> a==5, then assigning to b --> b == 5
int e = a++;  // Assigning a to e --> e==5, then incrementing a --> a==6

a *= b;       // Multiplying a with b and assigning the result to a --> a==30

System.out.println("a=" + a);
System.out.println("b=" + b);
System.out.println("c=" + c);
System.out.println("e=" + e);

exercise No. 54

Cleaning up the mess

Some developers follow the rule “It was hard to write, it should be hard to understand as well”. One such proponent codes:

int a = 3, b = 6, c = 2;

a += (a = 2) + ++a - ++b % c--;

System.out.println("a = " + a + ", b = " + b + ", c = " + c);

Execution results in:

a = 7, b = 7, c = 1

Decompose this cryptic assignment into a series of multiple elementary ones like e.g. a++ by possibly introducing one or more helper variables. Thus your code shall become longer but better to understand.

Tip

Read about evaluation of expressions and operator precedence.

We start by modifying the line in question:

a = a + (a = 2) + ++a - ++b % c--;

The expression on the right hand side will be decomposed by the Java™ compiler into a sum of four subexpressions to be evaluated from left to right:

a
a=2
++a
++b % c--

This one deserves further attention: The % operator will act on b's value after incrementing (infix notation) and c's value before decrementing (postfix notation).

We introduce a helper variable sum for decomposing our code into the above described four subexpressions:

int a = 3, b = 6, c = 2;

int sum = a;  // First subexpression
a = 2;
sum += a;     // Second subexpression
a++;
sum += a;     // Third subexpression
b++;
sum -= b % c; // Fourth subexpression
c--;

a = sum;      // Assigning the result

System.out.println("a = " + a + ", b = " + b + ", c = " + c);

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