Exercise 1.34. Suppose we define the procedure,only one ,are you joke

gyumni | Sept. 4, 2019, 3:04 a.m.

is not lambda is not let g is sqare g is a term so f(f2)

=22

you put term in it'self it will be never get stop

like y=x+3

but 

y y

second f is the (f g)

you put term in it self,

or mean do't  f g,but defind some use f n in the local name guess x

use x

term = square

you use sometime like term a and it like lambda

and 

square x x*x

square square

there never be stop  

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f f

f 2

2 2  

 f f1->f 22-> 2 23

1fg

g2

->f f

2

ff

-> ff

f2

iterative or rec

fg

->f 2

g2

22

 

3

f2

22

ff->22

g=2

f=2

bound variable g

something like if let x 3 x*x+2 x be the bound varible now f is

f = 2 

expn = g 2

f is x

g is 3 

3 2 is the value of f is let now is not l so

ff

f2

f2

22

ff

22

 

in the lambda 

x y

x+2 -y+2

x y

so  

in inter defines

define f x y

   de x

   de y

  (x+y)

 

 

 

> (f f)

. . application: not a procedure;

 expected a procedure that can be applied to arguments

  given: 2

  arguments...:

> (f square)

. . square: undefined;

 cannot reference an identifier before its definition

> (f +)

2

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