Lambda Calculus

Re:Lambda Calculus by btilly on Monday October 12, 2009 @11:44AM (#876947)

I could easily be wrong, but here is my understanding. Consider this snippet of JavaScript:

  function foo (x) { return function (y) {
      return x*y;
  }}

  bar = foo(5);

Within bar, x is bound and y is free.

Does that make sense to you?

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Re:Lambda Calculus by jacquesm on Monday October 12, 2009 @12:00PM (#876995)

Ok, so now 'bar' contains a function where the 'x' is already bound to the '5', but the 'y' is still free because it has not yet been 'bound' to any parameter, is that what you mean ?

bar contains "function(y) { return 5*y; }"

The 'x' has been replaced by the '5' because of the function call. So 'binding' effectively pins down a function parameter and allows you to replace its instances in 'named' form with its value to reduce the expression ?

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Re:Lambda Calculus by btilly on Monday October 12, 2009 @12:08PM (#877023)

That is the way that I understand it. However you should be warned that I have never made a detailed study of the lambda calculus, and so might have basic misconceptions about it.

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