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0 pointskazinator3y ago0 comments

If we don't have quote for expressions but only for symbols, we can still write the interpreter function, and give it a test case by writing an expression which calculates the code that we want to interpret. Even without quote the language has given us a representation of the syntax that we can rely on.

We have it as a given that (list 'lambda (list 'x) 'x) produces (lambda (x) x).

We do not have such a thing in lambda calculus. We could create an extended lambda calculus which has it.

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anderskaseorg3y ago

The sense in which the Lisp code (list 'lambda (list 'x) x) evaluates to the data representation of (lambda (x) x) is exactly the same as the sense in which the lambda calculus code λa. λb. λc. c (λx. λa. λb. λc. a x) evaluates to the data representation of λx. x.

What makes Lisp special is that this particular correspondence is made visible to the programmer via quote and other macros. Again, very cool, but not fundamental to this particular discussion.

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