In using sum as in section , it seems terribly awkward to have to define trivial procedures such as pi-term and pi-next just so we can use them as arguments to our higher-order procedure. Rather than define pi-next and pi-term, it would be more convenient to have a way to directly specify ``the procedure that returns its input incremented by 4'' and ``the procedure that returns the reciprocal of its input times its input plus 2.'' We can do this by introducing the special form lambda, which creates procedures. Using lambda we can describe what we want as
(lambda (x) (+ x 4))and
(lambda (x) (/ 1.0 (* x (+ x 2))))Then our pi-sum procedure can be expressed without defining any auxiliary procedures as
(define (pi-sum a b) (sum (lambda (x) (/ 1.0 (* x (+ x 2)))) a (lambda (x) (+ x 4)) b))
Again using lambda, we can write the integral procedure without having to define the auxiliary procedure add-dx:
(define (integral f a b dx) (* (sum f (+ a (/ dx 2.0)) (lambda (x) (+ x dx)) b) dx))
In general, lambda is used to create procedures in the same way as define, except that no name is specified for the procedure:
(lambda (formal-parameters) body)The resulting procedure is just as much a procedure as one that is created using define. The only difference is that it has not been associated with any name in the environment. In fact,
(define (plus4 x) (+ x 4))is equivalent to
(define plus4 (lambda (x) (+ x 4)))We can read a lambda expression as follows:
(lambda (x) (+ x 4)) the procedure of an argument x that adds x and 4
Like any expression that has a procedure as its value, a lambda expression can be used as the operator in a combination such as
((lambda (x y z) (+ x y (square z))) 1 2 3) 12or, more generally, in any context where we would normally use a procedure name.^{}
Using let to create local variables
Another use of lambda is in creating local variables. We often need local variables in our procedures other than those that have been bound as formal parameters. For example, suppose we wish to compute the function
which we could also express as
(define (f x y) (define (f-helper a b) (+ (* x (square a)) (* y b) (* a b))) (f-helper (+ 1 (* x y)) (- 1 y)))
Of course, we could use a lambda expression to specify an anonymous procedure for binding our local variables. The body of f then becomes a single call to that procedure:
(define (f x y) ((lambda (a b) (+ (* x (square a)) (* y b) (* a b))) (+ 1 (* x y)) (- 1 y)))This construct is so useful that there is a special form called let to make its use more convenient. Using let, the f procedure could be written as
(define (f x y) (let ((a (+ 1 (* x y))) (b (- 1 y))) (+ (* x (square a)) (* y b) (* a b))))The general form of a let expression is
(let ((var_{1} exp_{1}) (var_{2} exp_{2}) (var_{n} exp_{n})) body)which can be thought of as saying
let | var_{1} have the value exp_{1} and |
var_{2} have the value exp_{2} and | |
var_{n} have the value exp_{n} | |
in | body |
The first part of the let expression is a list of name-expression pairs. When the let is evaluated, each name is associated with the value of the corresponding expression. The body of the let is evaluated with these names bound as local variables. The way this happens is that the let expression is interpreted as an alternate syntax for
((lambda (var_{1} ...var_{n}) body) exp_{1} exp_{n})No new mechanism is required in the interpreter in order to provide local variables. A let expression is simply syntactic sugar for the underlying lambda application.
We can see from this equivalence that the scope of a variable specified by a let expression is the body of the let. This implies that:
(+ (let ((x 3)) (+ x (* x 10))) x)is 38. Here, the x in the body of the let is 3, so the value of the let expression is 33. On the other hand, the x that is the second argument to the outermost + is still 5.
The variables' values are computed outside the let. This matters when the expressions that provide the values for the local variables depend upon variables having the same names as the local variables themselves. For example, if the value of x is 2, the expression
(let ((x 3) (y (+ x 2))) (* x y))will have the value 12 because, inside the body of the let, x will be 3 and y will be 4 (which is the outer x plus 2).
Sometimes we can use internal definitions to get the same effect as with let. For example, we could have defined the procedure f above as
(define (f x y) (define a (+ 1 (* x y))) (define b (- 1 y)) (+ (* x (square a)) (* y b) (* a b)))We prefer, however, to use let in situations like this and to use internal define only for internal procedures. ^{}
Exercise. Suppose we define the procedure
(define (f g) (g 2))Then we have
(f square) 4What happens if we (perversely) ask the interpreter to evaluate the combination (f f)? Explain.(f (lambda (z) (* z (+ z 1)))) 6