Environments and the Contour Model

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Presentation transcript:

Environments and the Contour Model

Environments Name Value X 2 Y (lambda (x) (+ x 2)) Z “testing” … Consist of: frames chained together Name Value X 2 Y (lambda (x) (+ x 2)) Z “testing” … Frames are:

Global Environment Name Value + … Name Value X 2 Y Primitives Name Value + … Name Value X 2 Y (lambda (x) (+ x 2)) Z “testing” … Global user declarations

Function Evaluation > (Y 2) Name Value + … Name Value X 2 Y (lambda (x) (+ x 2)) Z “testing” … Name Value X 2 …

Function Evaluation So, X is bound in Y What about +? Unbound When unbound lookup …

Traverse Frames Name Value + … Name Value X 2 Y (lambda (x) (+ x 2)) Z “testing” … Name Value X 2 …

What about multiple bindings? (define x 3) (define y 2) (define z (lambda (x) (let ((y 3)) (+ x y)))) (z 4) X and Y multiply defined. Always use first binding! Name Value +  … Name Value X 3 Y 2 Z  Name Value X 4 Y 3

Passing Mechanism By value: Copy of argument is value for formal argument Pointer in the case of lists Copy of string or integer for primitive types

Primitive-Environment User-Environment Points at line where  defined b 5 square (2) Points at environment in which  defined line 1: (define b 5) line 2: (define (square num) line 3: (* num num)) line 4: (define result (square 5)) P = 4 Points at line where execution snapshot created

Points at line where function execution ends Primitive-Environment User-Environment b 5 square (2) square’ line 1: (define b 5) line 2: (define (square num) line 3: (* num num)) line 4: (define result (square 5)) num 5 P = 3 P = 4 Points at line where function execution ends Return pointer

Primitive-Environment User-Environment b 5 square 2 result 25 line 1: (define b 5) line 2: (define (square num) line 3: (* num num)) line 4: (define result (square 5)) P = 5

Primitive-Environment User-Environment factorial (1) line 1: (define (factorial n) line 2: (if (< n 2) line 3: 1 line 4: (* n (factorial (- n 1))))) line 5: (define result (factorial 3)) P = 5

Primitive-Environment User-Environment factorial (1) factorial’ n 3 line 1: (define (factorial n) line 2: (if (< n 2) line 3: 1 line 4: (* n (factorial (- n 1))))) line 5: (define result (factorial 3)) P = 4 P = 5

Primitive-Environment User-Environment factorial (1) factorial’ n 3 P = 4 P = 5 factorial’’ line 1: (define (factorial n) line 2: (if (< n 2) line 3: 1 line 4: (* n (factorial (- n 1))))) line 5: (define result (factorial 3)) n 2 P = 4

Primitive-Environment User-Environment factorial (1) factorial’ n 3 P = 4 P = 5 factorial’’ n line 1: (define (factorial n) line 2: (if (< n 2) line 3: 1 line 4: (* n (factorial (- n 1))))) line 5: (define result (factorial 3)) 2 P = 4 factorial’’’ n 1 P = 4

Primitive-Environment User-Environment factorial (1) result 6 P = 6

Warning Simplified Diagrams! Next slides are simplified: Lambdas for a, b, etc not present Extra environment frame is for the let inside of main. Remember: (let ((a 0) (b 0)) <body>) is equivalent to ((lambda (a b) <body>) 0 0) Main’ a 5 b 6 f (3) g (8)

Code Line 1: (define (main) Line 2: (let (( a 0) (b 0)) Line 3: (define (f) Line 4: (let ((a 0)) Line 5: (set! a b) Line 6: (writeln "in f: A and B are - " a b) Line 7: (g))) Line 8: (define (g) Line 9: (let ((b 0)) Line10: (set! b (+ a 2)) Line11: (writeln "in g: A and B are - " a b) Line12: (f))) Line13: (set! a 5) Line14: (set! b 6) Line15: (f))) Line16: (define a 10) Line17: (define b 20) Line18: (main)

Primitive-Environment User-Environment a 10 b 20 main (1) Main’ a 5 P = 15 b 6 f (3) g (8) F’ a 6 P = 7

Primitive-Environment User-Environment a 10 b 20 main (1) Main’ a 5 P = 15 b 6 f (3) g (8) G’ F’ a 6 b 7 P = 7 P = 12

Primitive-Environment User-Environment a 10 b 20 main (1) Main’ a 5 P = 15 b 6 f (3) g (8) F’ G’ F’’ a 6 b 7 a 6 P = 7 P = 12 P = 7

Primitive-Environment User-Environment a 10 b 20 main (1) Main’ a 5 P = 15 b 6 f (3) g (8) F’ G’ F’’ G’’ a 6 b 7 a 6 b 7 P = 7 P = 12 P = 7 P = 12

Primitive-Environment User-Environment a 10 P = 18 b 20 main (1) Main’ P = 15 F’ a 5 a 6 P = 7 b 6 f (3) g (8)

Primitive-Environment User-Environment a 10 P = 18 b 20 main (1) Main’ P = 15 F’ a 5 a 6 P = 7 b 6 G’ f (3) b 8 g (8) P = 12

Primitive-Environment User-Environment a 10 P = 18 b 20 main (1) Main’ F’ P = 15 a 5 a 6 P = 7 b 6 G’ f (3) b 8 F’’ g (8) a 8 P = 12 P = 7

Primitive-Environment User-Environment a 10 P = 18 b 20 main (1) Main’ F’ P = 15 a 5 a 6 P = 7 b 6 G’ f (3) b 8 F’’ g (8) G’’ a 8 b 10 P = 12 P = 7 P = 12