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55 lines
1006 B
55 lines
1006 B
2 years ago
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Exercise 3.12: The following procedure for
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appending lists was introduced in 2.2.1:
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(define (append x y)
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(if (null? x)
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y
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(cons (car x) (append (cdr x) y))))
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Append forms a new list by successively consing the elements of
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x onto y. The procedure append! is similar to
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append, but it is a mutator rather than a constructor. It appends the
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lists by splicing them together, modifying the final pair of x so that
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its cdr is now y. (It is an error to call append! with an
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empty x.)
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(define (append! x y)
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(set-cdr! (last-pair x) y)
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x)
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Here last-pair is a procedure that returns the last pair in its
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argument:
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(define (last-pair x)
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(if (null? (cdr x))
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x
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(last-pair (cdr x))))
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Consider the interaction
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(define x (list 'a 'b))
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(define y (list 'c 'd))
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(define z (append x y))
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z
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(a b c d)
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(cdr x)
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⟨response⟩
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(define w (append! x y))
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w
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(a b c d)
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(cdr x)
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⟨response⟩
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What are the missing ⟨response⟩s? Draw box-and-pointer diagrams to
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explain your answer.
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