1. Expressions creating information

2. topics  operators: precedence, associativity, parentheses, overloading  operands: side-effects, coercion  arithmetic, relational, boolean, assignment

3. Arithmetic  +, -, *, /, sometimes **  fully parenthesized expressions easy ((3*5)+((6/2)-(7+1)))  associativity and precedence for user’s convenience 3*5 + 6/2 -(7+1)  BUT no implicit operations: 2ab 3

4. Precedence **FORTRAN++ -- C++, java * /+ - unary + -* / % + - binary no precedence APL fully parenthesized pure LISP

5. Associativity (at precedence level)  left to right most common 3 + 4 – 5 + 6  8  right to left rare 2 ** 2 ** 3  256 ( ** FORTRAN) 3 X 4 + 2  18 ( no precedence APL )  no associativity ( ** Ada) 2 ** 2 ** 3  error (2 ** 2) ** 3  64 OR2 ** (2 ** 3)  256

6. The trinary expression  op1 ? op2 : op3 c, c++, java (3 + 4 < 10) ? 12 – 4 : 40 * 40  8  why () on op1? first expression is boolean

7. Operands  Side effects  Type conversions

8. Operands - side effects  Side effect - procedure changes a data value of a global variable or parameter var int d,g,h; begin function f(int v) d := 30; begin g := 20; v := 100; h := 10; g := 200; d := f(h) f := 300end. end; d is 300, h is 100, g is 200 {parameter v passed by reference}

9. Side effects and order of operand evaluation  side effects make order of operand evaluation important var int d,g,h; begin function f(int v) d := 30; begin g := 20; v := 100; h := 10; g := 200; d := g + f(h) f := 300end. end;Is d = 320 or 500?

10. Type conversion  what determines type of expression result? 1. operator if not overloaded: e.g. Pascal 3 div 4 is 0 but 3 / 4 is 0.75 2. operands if operator is overloaded: e.g. in java 3/4 is 0 but 3.0/4 is 0.75 coercion: implicit translation of value to another type standard type conversion rule - coercion to a ‘wider’ type is OK

11. Relational expressions  same precedence, no associativity operands are ‘numeric’ and result is boolean = or ==>>=.EQ..GT..GE. != or <><<=.NE. or /=.LT..LE.

12. Relational expressions  same precedence, no associativity operands are ‘numeric’ and result is boolean EXCEPT c – uses 0 for false, ~0 for true so result (1) is int and can be compared: 6 > 5 > 4  0 (false) why? 6 > 5  1 (true) then 1 > 4  0 (false)

13. Logical or boolean  boolean operands and result  operators unary (not) and binary (and, or,...) FORTRANAdac,c++,java.AND.andand then&&&.OR.oror else||| xor SHORT-CIRCUIT EVALUATION

14. Short Circuit Evaluation  stop evaluation if value of expression known (a * b) * ( 1 + b/a) stop if b==0? if a==0?  not used in arithmetic  CAN be used in logical expressions

15. Short Circuit Evaluation of Boolean expressions  op1 && op2 if op1 is false, expression value is false  op1 || op2 if op1 is true, expression value is true

16. Short Circuit example int i=0; while ( i<arr.length && arr[i]!= k ) i++; if k is not found, either short circuit or range error && and || short circuit in c, c++, java

17. Assignment 1.high level ‘store’ 2.binary operator 3.type matching of operands 4.return a value? i.e. assignment statement or assignment expression?

18. Assignment Target (left side)  memory cell (not data value like most operands)  type matching – strongly typed or typeless determines coercion of right side or left side  dynamic binding of cell to expression

19. Type matching (mixed mode)  typeless language a = 3.4 * y; // a becomes float a = “tttt”; // a becomes string  strongly type-checked language double a = 4 * 5 ; // int 20 coerced to double int a = 4.1 * 3.2; // type mismatch error

20. Assignment Target Binding static: x = 3 * y + 5; dynamic: a[i-4] = 3 * y + 5; ((Point)pt).xVal = 3 * y + 5; odd ? oddVale : evenVal = 3 * y + 5;

21. Assignment Orthogonality c, c++, java – operators combining arithmetic and assignment: not orthogonal unary ++: count++ binary +=: sum += count

22. Multiple target assignment  PL/1: multiple parallel targets A, B, C, D = 100;  c, c++, java: assignment expression returns value A = B = C = D = 100; right to left associativity

23. The assignment/expression mess in the legacy of c  goal: efficiency count++; sum += x; a = b = c = 0;  consequence: nested expressions and assignments – readability? a = b + (c = d / b++) – 1; (Sebesta, p.303)

24. The assignment/expression mess in the legacy of c a = b + (c = d / b++) – 1; (Sebesta, p.303) 24 1,3 5 67 ?