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Rahman Lavaee Mashhadi Mohammad Shadravan Conditional expressions LISP was the first language to contain a conditional expression In Fortran and Pascal we have to drop from expression level to statement level (cond (p1 e1) … (pn en)) ( if p e1 e2) = (cond (p e1) (t e2)) example :( defun sg(x) (cond((plusp x) 1) ((zerop x) 0) ((minusp x) -1)) ) Logical connectives Logical connectives are evaluated conditionally (or x y) = (if x t y) Their operands are evaluated sequentially (or (eq (car L) ‘key) (null L) ) wrong Strict interpretation vs. sequential interpretation Iteration Iteration is done by recursion Analogous to while-loop (defun plus-red (a) (if (null a) 0 (plus (car a) (plus-red (cdr a)) )) ) There is no “bounds” , no “explicit indexing” , no “control variable” It obeys the zero-one-infinity principle Nested Loops Example : Cartesian product (defun all-pairs (M N) (if (null M) nil (append (distl (car M) N) (all-pairs (cdr M ) N )) )) (defun distl (x N) (if (null N) nil (cons (list x (car N)) (distl x (cdr N)) )) ) Hierarchical structures Are difficult to handle iteratively example: equal function eq only handles atoms initial states If x and y are both atoms (equal x y) = (eq x y) If exactly one of x and y is atom (equal x y) = nil (and (atom x) (atom y) (eq x y)) use car and cdr to write equal recursively Equivalency of recursion and iteration it may be seemed that recursion is more powerful than iteration in theory these are equivalent As we said iteration can be done by recursion by maintaining a stack of activation records we can convert a recursive program to an iterative one. Functional arguments and abstraction Suppress details of loop control and recursion example: applying a function to all elements of list (defun mapcar (f x) (if (null x) nil (cons (f (car x)) (mapcar f (cdr x)) )) ) (defun reduce (f a x) (if (null x) a (f (car x) (reduce f a (cdr x) )) ) ) Functional arguments and combination of programs Functional arguments simplify the combination of already implemented programs example : inner product of two lists (defun ip (u v) (reduce ‘plus 0 (mapcar2 ‘times u v)) ) Anonymous functions It is inconvenient to give a name to every function we want to pass it only once (defun consval (x) (cons val x)) An obvious solution is just to pass the function’s body (mapcar ‘(cons val x) L) It is ambiguous : we do not know which names are parameters and which ones are global “lambda expressions” come in hand (mapcar ‘(lambda (x) (cons val x)) L) Functionals A functional is a function that has either(or both) of the following: One or more functions as arguments A function as its result Examples: Mapcar, reduce ,… Converting a binary function to a unary one Functional can replace lambda expressions Specifying parameters of a function with lambda expressions can be done by functionals example: (lambda (y) (f x y)) define “bu” function to fix one parameter of “f” (defun bu (f x) (function (lambda (y) (f x y)) )) (lambda (y) (f x y)) ~ (bu f x) (lambda (x y) (f y x)) ~ (rev f) Combining functions Flexibility of combining functions can be increased by using the uniform style. Take a function , and return a function ((map ‘add1) L) = (mapcar ‘add1 L) This form is called ‘combining forms’ Combining existing programs to acomplish new tasks example: (set ‘vec-dbl (map (bu ‘times 2)) ) (vec-dbl ‘(1 4 3 2)) (2 8 6 4) Backus style for functional programming alternate functional style of programming Based on ‘combining form’ Programming at a higher level of abstraction Simple algebraic manipulation combining, reversing, mapping functions Function-level vs. object-level(manipulate function rather objects) Backus notation Name Lisp Backus Application Mapping Reduction Composition Binding Constant Lists Built-in-functions Selectors (f x) (map f) (red f a) (comp f g) (bu f k) (const k) (a b c d) plus, times, … cdr, car, cadr, … f:x αf /f fog (bu f k) k© (a,b,c,d) +,×,… tail, 1 , 2 Variable-free programming Variables (i.e. formal parameters and algol scope rules) complicate manipulating programs One of the goals of this notation was to eliminate variables Elimination of lambda expressions by using bu and rev functionals Example: inner product function Trans:<<…ui…> , <…vi…>> = <….<ui,vi>…> (α ×): (trans: <u,v>)=<u1v1,…,unvn> Ip: <u,v> = (/+): ((α ×): (trans: <u,v>)) Variable elimination Ip= (/+)o (α ×)o trans Name structures Value bindings are established in two ways: Property lists Established by pseudo functions such as ‘set’ and ‘defun’ example: (set ‘text ‘(to be or not to be)) They are global low level execution with property lists example: (putprop ‘add1 ‘(lambda (x) (+ x 1)) ‘expr) Actual formal correspondence Like argument passing in other languages example : (defun times (x y) (* x y)) (times 3 FACT) the formal x will be bound to atom 3 and the formal y will be bound to the value of actual FACT Temporary binding Binding names in a local context like in algol’s blocks is done by actual Prevents duplicate calculation Obeys abstraction principle (defun roots (a b c) (list (/ (+ (- b) (sqrt (- (expt b 2) (* 4 a c)) )) (* 2 a)) (/ (+ (- b) (sqrt (- (expt b 2) (* 4 a c)) )) (* 2 a)) )) solution (defun roots-aux (d) (list (/ (+ (- b) d) (* 2 a)) (/ (+ (- b) d) (* 2 a)) )) (defun roots (a b c) roots-aux (sqrt (- ( expt b 2) (times 4 a c)) )) ) Let function Function calling with let (let ((n1 e1) … (nm em)) E) Abbreviation for a function definition and application (defun QQQQQ (n1 … nm) E) (QQQQQ e1 …. em) Allows a number of local names to be bound to values at one time The second argument is evaluated in the resulting environment and is returned as the value of let Like alogol block , Ada declare block Dynamic scoping review Dynamic scoping: functions are called in the environment of its caller Static scoping: functions are called in the environment of its definition LISP uses dynamic scoping example : roots-aux had access to the names ‘a’ and ‘b’ The ‘let’ function needs dynamic scoping Dynamic scoping complicates functional arguments (defun twice (func val) (func (func val)) ) (twice ‘add1 5) returns 7 (twice ‘(lambda (x) (times 2 x)) 3) returns 12 func val DL 3 x DL 3 a (times 2 x) Functional argument problem (set ‘val 2) 2 (twice ‘(lambda (x) (times val x)) 3) What is the return value of this expression? we expect it to be 12 but it will return 27! why? Cause :Collision of variables val 2 func val DL 3 x 3 DL (times val x) What’s the solution? What should we do? use different names In a large program it would be impractical to ensure that all of the names are distinct change the scoping rule ‘function’ primitive Use ‘function’ primitive to bind the lambda expression to its environment of definition (twice (function (lambda (x) (times val x)) ) 3 ) this will return 12 val 2 func val x 3 3 (times val x) Syntactic structures Convenience for programmers is not the main intention of the S-Expression ‘cond’ syntax was improved for convenience of programmers (cond (p1 e1) (p2 e2) … (pn en)) (cond p1 e1 p2 e2 … pn en) ‘define’ function was used for global binding (define ((n1 e1) … (nk ek)) ) Syntactic structures (define ( (add1 (lambda (x) (+ x 1)) )) ) Lots of idiotic single parentheses ‘defun’ was declared (defun (add1 x) (+ x 1)) (quote val) was improved to ‘val (set (qutoe val) 2) was improved (setq val 2) List representation of programs Facilitates program manipulation’ Balances reduction of readability Simple to write LISP programs that generate other LISP programs Separating parts of a function (car A) is the name of the function A (cdr A) is the arguments of the function A Obeys Elegance Principle : writing LISP interpreter in LISP