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Writing functions in OCaml

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Defining a simple function # let add (x, y) = x + y;; val add : int * int -> int = Notice what this says: –add is a value –the type of add is int * int -> int –the -> denotes a function –Therefore, a function is a kind of value! –The actual value is abbreviated to

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Another function definition # let double x = x + x;; val double : int -> int = Why don't we need parentheses around the parameter x ? Every function in OCaml takes one parameter and returns one result. –The parameter may be a tuple –The result may be a tuple

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add again # let add (x, y) = x + y;; val add : int * int -> int = # add (7, 3);; - : int = 10 # add (3.0, 5.0);; Characters 5-13: This expression has type float * float but is here used with type int * int OCaml is strongly typed, but it can deduce types

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Statements in OCaml There are no statements in OCaml OCaml is (almost) a purely functional language; it has no side effects* –but OCaml does have expressions Every expression has a value * Except for output "statements"--we won't be doing output

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The if...then...else expression if boolean-expression then expression 1 else expression 2 The else part is required (why?) –because the if expression must have a value expression 1 and expression 2 must have the same type –because OCaml is strongly typed –it needs to know the type of the expression

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Using if...then...else # let max (x, y) = if x > y then x else y;; val max : 'a * 'a -> 'a = # max (7, 5);; - : int = 7 # max (7.0, 5.0);; - : float = 7.000000 max, as defined here, is a polymorphic function

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Integer division, with remainder # let divide (x, y) = x / y, x mod y;; val divide : int * int -> int * int = # divide (20, 3);; - : int * int = 6, 2 We aren't returning two results, just one--but it's a tuple Similarly, we're only providing one parameter

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Adding vectors # let addVec ((x1, y1), (x2, y2)) = (x1 + x2, y1 + y2);; val addVec : (int * int) * (int * int) -> int * int = # addVec ((3, 5), (10, 20));; - : int * int = 13, 25 Notice that parentheses are sometimes necessary

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LISP-like operations Recall that OCaml has the following operations: –hd returns the head of a list –tl returns the tail of a list –:: adds an element to a list These are essentially the same as CAR, CDR, and CONS in LISP Can we define CAR, CDR, and CONS in OCaml?

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Redefining LISP # let car x = List.hd x;; val car : 'a list -> 'a = # let cdr x = List.tl x;; val cdr : 'a list -> 'a list = # let cons (x, y) = x :: y;; val cons : 'a * 'a list -> 'a list = # car (cdr [1; 2; 3; 4]);; - : int = 2

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Testing our LISP functions cons ([1; 2], [3; 4]);; # Characters 6-20: This expression has type int list * int list but is here used with type int list * int list list Elements of list must be same type! There is no solution (with the OCaml we know so far)

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Trapped in time # let age = 20;; val age : int = 20 # let older () = age + 1;; val older : unit -> int = # older ();; - : int = 21 # let age = 35;; val age : int = 35 # older ();; - : int = 21

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OCaml adds to state, doesn't change it OCaml has no "assignment" let age = 21; associates age with 21 This age is then used in function older let age = 35; associates a new age with 35 The old age goes out of scope, but it doesn't go away (it's lifetime is not over) older still uses the original variable

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Functions are values, too # let f x = 2 * x;; val f : int -> int = # let g x = f (f x);; val g : int -> int = # g 5;; - : int = 20 # let f x = 3 * x;; val f : int -> int = # g 5;; - : int = 20

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Debugging functions You saw that redefining a function does not affect prior uses of that function Therefore, you can't change a function by changing something it calls Everything should work OK if you compile all your functions from a file every time Sometimes you just have to restart OCaml

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It's all in the binding times OCaml's approach seems strange, but you have seen it before x = 5; y = 2 * x; x = x + 1; print "y = ", y; Would you expect y = 12 to be printed? Remember, functions are values too!

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The End

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