CSE305 – Programming Languages Daniel R. Schlegel April 25, 2011

CSE305 – Programming LanguagesDaniel R. SchlegelApril 25, 2011 “Should array indices start at 0 or 1? My compromise of 0.5 was rejected without, I thought, proper consideration.”(Stan Kelly-Bootle)

JavaScript Syntax • Is this valid? int fact(int n) { if (n === 0) { return 1; } else return n * fact(n - 1); }

JavaScript Syntax • Is this valid? int fact(int n) { if (n === 0) { return 1; } else return n * fact(n - 1); } NO! function fact(n) { if (n === 0) { return 1; } else return n * fact(n - 1); }

JavaScript Syntax • What does this return? function array(f, g) { //f and g are functions return [f(), g()]; }

JavaScript Syntax • What does this return? • What about this? function array(f, g) { //f and g are functions return [f(), g()]; } function array(f, g) { //f and g are functions return [f, g]; }

Assignment 5, Part 1 iteratorireverse(array a) { int n = a.length -1; for (inti=0; i<=n; i++) { yield a[n-i]; } } void main { foreach x inireverse(a) { foreach y inireverse (b) { sum = sum + x*y; } } } Q: When you call ireverse(a) does it give you all the answers at once?

Assignment 5, Part 1 iteratorireverse(array a) { int n = a.length -1; for (inti=0; i<=n; i++) { yield a[n-i]; } } void main { foreach x inireverse(a) { foreach y inireverse (b) { sum = sum + x*y; } } } Q: When you call ireverse(a) does it give you all the answers at once? NO! What does this do?

Assignment 5, Part 1 iteratorireverse(array a) { int n = a.length -1; for (inti=0; i<=n; i++) { yield a[n-i]; } } void main { foreach x inireverse(a) { foreach y inireverse (b) { sum = sum + x*y; } } } Q: When you call ireverse(a) does it give you all the answers at once? NO! What does this do? Get the next value of ireverse(a). Set x to it.

We need to simulate this behavior! • void main(){ • List l1, l2; … • l2 := nil; • foreach x in elements(l1) do{ • l2 := cons(x, l2); • } • print l2; • } • iteratorint elements(List l){ • while(l != nil) { • yield l.val; • l := l.next; • } • } • void main(){ • List l1, l2; • void thk(int x) { • l2 := cons(x, l2); • } • l2 := nil; • elements(l1, thk); • print l2; • } • void elements(List l, void thk(int)){ • while (l != nil) do { • thk(l.val); • l := l.next; • } • }

Lazy Evaluation • What’s the advantage?

Lazy Evaluation • What’s the advantage? • Doesn’t find a result until that result is needed! • We can use functions that generate infinite lists. • Does JavaScript have it built in?

Lazy Evaluation • What’s the advantage? • Doesn’t find a result until that result is needed! • We can use functions that generate infinite lists. • Does JavaScript have it built in? • No! • Can we simulate it? If yes, how?

Lazy Evaluation • What’s the advantage? • Doesn’t find a result until that result is needed! • We can use functions that generate infinite lists. • Does JavaScript have it built in? • No! • Can we simulate it? If yes, how? • Yes, using higher order functions.

How we usually do things… function f(x) { return [x+1, x*2, x*x]; } f(100); //[101, 200, 10000] Short running tasks Okay, so these are short running tasks… but what if we didn’t need all of this?

What about this… function f(x) { return [longtask(x), longertask(x), reallylongtask(x)]; } function g(n) { print(f(100)[n]); //Some value } Is there a problem with this? We never use f(100)[0] or f(100)[2], but they are still evaluated!

A lazy version function f(x) { thk1() = longtask(x); thk2() = longertask(x); thk3() = reallylongtask(x); return [thk1, thk2, thk3]; } function g(n) { print(f(100)[n]()); //Some value }

So now extend this to infinite lists… numsfrom = function(n){ return cons(n, numsfrom(n+1)); }; What happens when we run this?

So now extend this to infinite lists… numsfrom = function(n){ varthk = function(){return numsfrom(n+1);}; return [n,thk]; };

Final Notes • JIVE Survey • Today is the last recitation  • Good luck on the assignment and final!  • Also, I’m teaching CSE111 this summer… If you know anyone who needs to take it, let them know!

CSE305 – Programming Languages Daniel R. Schlegel April 25, 2011