1. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Java Operators Recall Java’s programming components: Packages - Collection of classes (Programs) Classes - Collections of data and methods that operate on data Methods - Collections of statements that operate on data Variables - Store data Constants - Store data that doesn’t change Literals - Explicit data like “Hello World”

2. ***** SWTJC STEM ***** Expressions and Operators Expressions are segments of code that perform computations and return values. They are combinations of literals, variables, constants, and operators. An operator performs an action on variables, constants, and literals. Associated with an operator are one or two operands that receive the action of the operator. Unary operators act on only one operand. Example: -12, negation operator “-” Binary operators act on two operands. Example: 8 + 15, addition operator “+” Chapter 2-3 cg 29

3. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Unary Operators Unary operators support either prefix or postfix notation. Prefix notation means that the operator appears before its operand: operator operand Example:  - i, negation operator “-”  ++ i, shortcut increment operator “++“ Postfix notation means that the operator appears after its operand: operand operator Example:  i ++, shortcut increment operator “++“

4. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Binary Operators Binary operators use infix notation, which means that the operator appears between its operands: operand1 operator operand2 Examples: i + j a / b x += 10

5. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Basic Numeric Operators OperatorUseDescription ++op Promotes op to int if it's a byte, short, or char --opArithmetically negates op Unary Operators (op is an operand) Operator + - * / % Use op1 + op2 op1 - op2 op1 * op2 op1 / op2 op1 % op2 Description Adds op1 and op2 Subtracts op2 from op1 Multiplies op1 by op2 Divides op1 by op2 Computes the remainder of dividing op1 by op2 Binary Operators

6. ***** SWTJC STEM ***** Chapter 2-3 cg 30 Numeric Operator Examples Declare Variables... int i = 37; int j = 42; double x = 27.475; double y = 7.22; Adding... i + j is 79 x + y is 34.695 Subtracting... i - j is -5 x - y is 20.255 x - (-y) is 34.695 Multiplying... i * j is 1554 x * y is 198.3695 Dividing... i / j is 0 (integer division implied) x / y is 3.8054... Computing the remainder... i % j is 37, j % i is 5 x % y is 5.815 Mixing types...converts to double... j + y is 49.22 i * x is 1016.58

7. ***** SWTJC STEM ***** Chapter 2-3 cg 30 Coding an expression - Examples Example: Coding expressions 1.int thisDecade = 2000; byte decadeLength = 10; int nextDecade = thisDecade + decadeLength;  byte decadeLength = 10; assigns the value 10 to a 8-bit byte variable named decadeLength.  int thisDecade = 2000; assigns the value 2000 to a 32-bit integer variable named thisDecade.  int nextDecade = thisDecade + decadeLength; assigns the value 2010 to a 32-bit integer variable named nextDecade.

8. ***** SWTJC STEM ***** Chapter 2-3 cg 30 Coding an expression - Examples Examples: Expression coding continued... 2.float radius = 2.0f; float area = 3.14159f * radius * radius;  float radius = 2.0f; assigns the value 2.0 to a 32-bit (single precision) floating point variable named radius.  float area = 3.14159f * radius * radius; assigns the computed expression to a 32-bit (single precision) floating point variable named area.

9. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Shortcut Incr/Decr Operators Shortcut Increment/Decrement Operators OperatorUseDescription ++ (incr)op++ Increments operand op by 1; evaluates to the value of op before incrementing op ++ (incr)++op Increments operand op by 1; evaluates to the value of op after incrementing op -- (decr)op-- Decrements operand op by 1; evaluates to the value of op before decrementing op -- (decr)--op Decrements operand op by 1; evaluates to the value of op after decrementing op

10. Note: --i and i– (decrement) work similarly. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Shortcut Incr/Decr Examples inewNum Before10Undeclared After11100 Postfix Operator Code... int i = 10; int newNum = 10 * (i++); i increments after *. Prefix Operator Code... int i = 10; int newNum = 10 * (++i); inewNum Before10Undeclared After11110 i increments before *.

11. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Shortcut Assignments Operators OperatorUseEquivalent to +=op1 += op2op1 = op1 + op2 -=op1 -= op2op1 = op1 - op2 *=op1 *= op2op1 = op1 * op2 /=op1 /= op2op1 = op1 / op2 %=op1 %= op2op1 = op1 % op2 Examples... iValue += 8; is the same as iValue = iValue + 8; jValue %= 5; is the same as jValue = jValue % 5;

12. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Shortcut Assignment Examples Assignment Addition Code: int i = 10; int newNum = 10; newNum += i; Same as newNum = newNum + i; newNum becomes 20. i remains 10. Assignment Division Code: int i = 10; int newNum = 10; newNum /= i; Same as newNum = newNum / i; newNum becomes 1. i remains 10.

13. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Character Operator Operator UseDescription + char1 + char2 string1 + string 2 Concatenates characters and strings Binary Operator A char/string variable holds character/string data, respectively. Char literals use apostrophes ‘s‘ and are single values. String literals use quotation marks “red leaf” and can hold more than one character (i.e., text). Concatenation “ties together” char/string variables and literals.

14. ***** SWTJC STEM ***** Chapter 2-3 cg 29 Character Operator Examples Examples... char cValue = ‘s’; Declares a character variable cValue and assigns character s to it. String c1Value = “red”, c2Value = “bird”; Declares string variables c1Value and c2Value assigning strings “red” and “bird” to them respectively. System.out.println(c1Value + c2Value + cValue); Outputs string redbirds to the PC monitor, which is the concatenation of all three variables.

15. ***** SWTJC STEM ***** Chapter 2-3 cg 29-31 Creating Complex Assignments When creating complex assignment expressions, 1.How is the order of operations determined? 2.What if we mix data and variable types? Examples: int result = 14 + 8 / 2;  Is the result 11 (+ first)? or 18 (/ first)? int result = 12 / 2 * 3;  Is the result 18 (/ first)? or 2 (* first)? float total = 100.46f; int count = 10; float result = total / count; // Mixed types?  Is result 10 (integer /)? or 10.046 (float /)?

16. ***** SWTJC STEM ***** Chapter 2-3 cg 29-31 Order of Operations - Precedence Expressions are evaluated according to operator precedence. When operators of equal precedence occur together in an expression, association governs the order. Left-associative means operations are performed left-to- right. Binary operators, except the assignment operator, are left-associative. Right-associative means operations are performed right- to-left. Unary and assignment operators are right- associative.

17. ***** SWTJC STEM ***** Chapter 2-3 cg 29-31 Order of Operations - Precedence Precedence Level OperatorOperationAssociates 1+ - ++,-- unary plus unary minus shortcut R to L 2*/%*/% multiplication division modulus L to R 3+-++-+ addition subtraction concatenation L to R 4= +=,-=,... assignmentR to L

18. ***** SWTJC STEM ***** Chapter 2-3 cg 29-31 Order of Operations - Examples Given differing orders of precedence. result = 14 + 8 / 2; // Divide first / higher precedence than + and result is 18. Precedence can be forced using parentheses. result = (14 + 8) / 2; // Add first + is forced first by parentheses and result is 11. Given the same order of precedence. result = 12 / 2 * 3; // Divide first / is first (L to R), then * and result is 18. Adding a unary operator -. result = 12 / -(-3 + 1) * 3; // Negation first - is first (R to L), then / (L to R), then * and result is 18.

19. ***** SWTJC STEM ***** Chapter 2-3 cg 29-31 Order of Operations - Assignment Given assignment operators. Assume int a = 1. 1.result = a + (a = 3); 1)Proceed L to R a = 1, then 1 + (a = 3). 2)Now do (a = 3) then 1 + 3 result = 4. 2.result = (a = 3) + a; 1)Proceed L to R (a = 3) first, so a is now 3. 2)Now do a + a = 3 + 3 so result is 6. 3.result = (a += 2 * a) + a 1)(a = 1 + 2 * 1) + a. 2)(a = 3) + a = a + a = 3 + 3 = 6.

20. ***** SWTJC STEM ***** Chapter 2-3 cg 29-31 Order of Operations - Incr/Decr Given increment/decrement operators. Assume int a = 5. 1.result = a + (--a) + a 1)Proceed L to R a = 5, then 5 + (--a) + a. 2)Now do a = a - 1= 4  a, then 5 + 4 + a  9 + a. 3)Finally, a = 4 and 9 + 4  result is 13. 2.result = a + (a--) + a 1)Proceed L to R a = 5, then 5 + (a--) + a. 2)Now do 5 + 5 + a = 10 + a, then do a = a - 1 = 4  a. 3)Finally 10 + a = 10 + 4  result is 14.

21. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Mixed Data/Variable Types Java is a strong typed language, meaning that all data and variables must be explicitly typed. Sometimes conversion is necessary. Recall float total = 100.46f; int count = 10; float result = total / count; // Mixed types? Care must be exercised or information can be lost. byte a = 20; // 1 byte in size int b = 2 * a ; // 4 bytes in size b can hold a; (1 byte) will fit in (4 bytes). int a = 2000; // 4 bytes in size byte b = 2 * a ; // 1 byte in size b cannot store a. ; (4 bytes) will not fit in (1 byte)!

22. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Mixed Data/Variable Types Conversions are of two types: 1.Widening conversions. Safest, does not lose information. Data conversion involves smaller to larger storage. doublefloat float, doublelong* long, float, doubleint int, long, float, doublechar int, long, float, doubleshort short, int, long, float, doublebyte ToFrom * Some precision (significant digits) could be lost!

23. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Mixed Data/Variable Types Conversions are of two types (continued): 2.Narrowing conversions. Likely to lose information. Avoid! Data conversion involves larger to smaller storage. byte, short, char, int, longfloat byte, short, char, intlong byte, short, charint byte, shortchar byte, charshort charbyte ToFrom byte, short, char, int, long, floatdouble

24. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Assignment Conversion Java conversions occur in three ways: 1.Assignment conversion Occurs when one type is assigned to another. Only widening conversions allowed. Is automatic. double money; int dollars = 1000; money = dollars; // int will fit in double OK - Widening conversion, done. double money = 1000; byte dollars; dollars = money; // double will not fit in int (max 255)! Not OK - Narrowing conversion, error!

25. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Arithmetic Promotion Java conversion occur in three ways (cont.): 2.Arithmetic promotion Occurs in expressions when operators need to widen (promote) data to complete. Only widening conversions allowed. Promotion is automatic when possible. float total = 100.46f, result; int count = 10; result = total / count; count promoted to float automatically. Floating point division used to calculate result. result is 1.0046.

26. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Casting - Forced Conversion Java conversion can occur in three ways: 3.Casting - Forcing the conversion Most general form of conversion. Unary operator specified by a data type in parentheses, (type) operand. Is not automatic. Programmer must specify. Example: float money = 8.75f; int dollars; dollars = (int) money; Casts money (float) to dollars (int). dollars is 8 money truncated (decimal portion dropped)

27. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Casting Continued Java conversion can occur in three ways: 3.(cont) Casting examples double result; int count = 10, total = 20; result = (double) total / count; total cast to double. count automatically promoted to double. Floating point division used. Quotient 2.0 stored in result as a double.

28. ***** SWTJC STEM ***** Chapter 2-3 cg 32 Casting Continued Java conversion can occur in three ways: 3.(cont) Casting char variables char c = 'A'; // upper case A int ic; ic = (int) c; char c is cast as an int ic is 65. ic = 97; c = (char) ic; int ic is cast as char c = ‘a’, lower case a.