Literals

1. Literals Integer literals: • Octal: 010, 027L, 017777777777 • Hexadecimal: 0x8, 0xaL, 0x7fffffff • Binary: 0b1000, 0b101010L • Decimal: 8, 64, 2147483647

2. Literals String literals: • “A” • “Hello” • “This is a good day for learning java.”

3. Literals float x = 2.5; • Why does this produce an error?

4. Literals float x = 2.5; • Floating-point default datatype is ‘double’.

5. Literals float x = (float) 2.5; • Solution 1: cast 2.5 to float

6. Literals float x = 2.5f; • Solution 2: use ‘f’ or ‘F’ identifier

7. Literals char c = ‘A’; • char literals are delimited by single quotes

8. Literals char c = ‘A’; int x = 45 + c; System.out.println( x );

9. numeric boolean integral floatingpoint character integer boolean char byte short int long float double 16 + 8 +/- 16 +/- 32 +/- 64 +/- 32 +/- 64 +/-

10. Literals char c = ‘A’; int x = 45 + c; System.out.println( x ); • char is a numeric integral type (i.e. evaluates to an integer • What is the output?

11. Literals • char c = ‘A’; char c = ‘\u0061’; • Unicode escape sequence is backslash + u + 4 digit hex • Unicode can be used ANYWHERE in java code

12. Literals ch\u0061r c = ‘\u0061’;

14. Numeric Promotion int x = 3 / 2 * 4 • 3,2 and 4 are int literals • Integer Division! • Result of 3 / 2 is : 1 • Therefore, x=3/2*4 is the same as x=1*4 • x = 4 !

15. Numeric Promotion int x = 3 / 2 * 4 f

16. Numeric Promotion int x = 3 / 2 * 4 f • The ‘f’ denotes ‘float’ datatype

17. Numeric Promotion int x = 3 / 2 * 4 f • The ‘f’ denotes ‘float’ datatype • The binary operator ‘/’ must have same datatype on both sides. Therefore, 2 is promoted to float datatype

18. Numeric Promotion int x = 3 / 2 * 4 f • The ‘f’ denotes ‘float’ datatype • The binary operator ‘/’ must have same datatype on both sides. Therefore, 2 is promoted to float datatype • Now the binary operator ‘*’ must promote its right hand operand. 4 is promoted to float.

19. Numeric Promotion int x = 3 / 2 * 4 f • The right hand expression evaluates to 6.0 • This will not compile because float to integer is a narrowing conversion requiring an explicit cast

20. Numeric Promotion int x = (int) 3 / 2 * 4 f • Explicit cast ! • System.out.print (x) : 4

21. Numeric Promotion float x = 3 / 2 * 4 f • cast not needed • System.out.print (x) : 6.0

22. Numeric Promotion float x = 3 / 2 * 4 f • Another method!

23. Numeric Promotion float x = 3.0 / 2 * 4 • Another method! • Literals are promoted to which datatype???

24. Numeric Promotion float x = 3.0 / 2 * 4 • Another method! • Literals are promoted to ‘double’ datatype

25. Numeric Promotion float x = 3.0 / 2 * 4 • Another method! • Literals are promoted to ‘double’ datatype • This expression will not compile because it requires an explicit narrowing conversion from double to float

26. Numeric Promotion float x = (float) 3.0 / 2 * 4 • Finished!

27. Numeric Promotion float x = (float) 3.0 / 2.0 * 4 • Does not compile.. Cast only operates on left hand operand 3.0

28. Numeric Promotion float x = (float) (3.0 / 2.0) * 4 • Phew!

29. Finally ! double x = 3 / 2 * 4.0 • evaluates left to right • 3 / 2 is integer division, result is 1 • Int 1 is promoted to double 1.0 • 1.0 * 4.0 evaluates to double 4.0

30. Literals long x = 2147483647; • What datatype is the literal?

31. Literals long x = 2147483648; • Why does the above produce an error? • By default, integer literals are datatype ‘int’, and 2,147,483,648 is out of range.

32. Literals long x = 2147483648L; • Solution: must use the ‘l’ or ‘L’ identifier

33. Literals int x = 2147483647 + 1; System.out.println( x ); • What is the output?

34. Literals 0 1111111 11111111 11111111 11111111 Sign bit: +/- • Int datatype is 32 bits: 1 sign bit + 31 bits to represent the number • Sign bit: 0 means positive, 1 means negative

35. Literals 0 1111111 11111111 11111111 11111111 + 0 0000000 00000000 00000000 00000001 = 1 0000000 00000000 00000000 00000000 Sign bit is now 1

36. Literals 2,147,483,647 0 1111111 11111111 11111111 11111111 + 0 0000000 00000000 00000000 00000001 = 1 0000000 00000000 00000000 00000000 1 - 2,147,483,648

37. Implicit Narrowing Conversion 0 1000000 10000001 • What is the decimal value of this bit pattern?

38. Implicit Narrowing Conversion 0 1000000 10000001 • 16513 ( 16384 + 128 + 1 )

39. Implicit Narrowing Conversion 0 1000000 10000001 short x = 16513; ‘16513’ is an integer literal. Default type is ‘int’. No cast required here because integral types (including ‘char’) can perform ‘implicit narrowing conversion’ as long as the value is in range. ( float c = 2.5 is illegal, because floating point types cannot perform implicit narrowing conversion.)

40. Implicit Narrowing Conversion • Rule 1: Both datatypes must be integer types • Rule 2: Value must be within range of target datatype

41. loss of precision with casting 0 1000000 10000001 short x = 16513; byte b = (byte) x;

42. loss of precision with casting short x = 16513; byte b = (byte) x; 0 1000000 10000001 bye, bye….

43. loss of precision with casting short x = 16513; byte b = (byte) x; 0 1000000 10000001 Value of x is -127

44. Exercise! int x = 457; double y = (byte) x; System.out.print( y );