From: Jim Pryor
Date: Thu, 25 Nov 2010 16:34:38 +0000 (-0500)
Subject: tweak calc improvements
X-Git-Url: http://lambda.jimpryor.net/git/gitweb.cgi?p=lambda.git;a=commitdiff_plain;h=88f7b0444146d9379521cc31ad9ef87b909e0b3b
tweak calc improvements
Signed-off-by: Jim Pryor
---
diff --git a/advanced_topics/calculator_improvements.mdwn b/advanced_topics/calculator_improvements.mdwn
index e22afb1a..4f24f877 100644
--- a/advanced_topics/calculator_improvements.mdwn
+++ b/advanced_topics/calculator_improvements.mdwn
@@ -120,7 +120,7 @@ We can begin with our language:
| Lambda of (char * term)
| Apply of (term * term);;
-Next, we need to expand our stock of `expressed_value`s to include function values as well. How should we think of these? We've several times mentioned the issue of how to handle free variables in a function's body, like the `x` in `lambda y -> y + x`. We'll follow the usual functional programming standard for these (known as "lexical scoping"), which keeps track of what value `x` has in the function expression's lexical environment. That shouldn't get shadowed by any different value `x` may have when the function value is later applied. So:
+Next, we need to expand our stock of `expressed_value`s to include function values as well. How should we think of these? We've several times mentioned the issue of how to handle free variables in a function's body, like the `x` in `lambda y -> y + x`. We'll follow the usual functional programming standard for these (known as "lexical scoping"), which keeps track of what value `x` has in the function declaration's lexical environment. That shouldn't get shadowed by any different value `x` may have when the function value is later applied. So:
let x = 1 in let f = lambda y -> y + x in let x = 2 in apply f 2
@@ -157,7 +157,7 @@ Now our evaluation function needs two further clauses to interpret the two new e
There are different ways to include recursion in our calculator. First, let's imagine our language expanded like this:
- let x = 1 in letrec f = lambda y -> if iszero y then x else y * f (y - 1) in f 3
+ let x = 1 in letrec f = lambda y -> if iszero y then x else y * apply f (y - 1) in apply f 3
where the AST would be: