为何列表元素比较时eq?返回false，其他场景却返回true？

Ah, I see the issue here—you've got a small but tricky syntax mistake that's throwing off the comparison! Let's break this down step by step.

First, let's clarify what '('leet 'a 'f) actually represents in Racket. When you write 'x, that's shorthand for (quote x). So the outer ' quotes the entire expression inside its parentheses. That means '('leet 'a 'f) expands to a list where each element is itself a quote expression—so the first element isn't the symbol leet, it's the list (quote leet) (which Racket displays as 'leet).

When you call (first '('leet 'a 'f)), you're getting this (quote leet) list object, not the plain symbol leet. Similarly, (first '('leet 'coder 'a 'f 'f)) returns another separate (quote leet) list object.

Remember that eq? checks for identity—it only returns #t if the two values are the exact same object in memory. Symbols are interned in Racket, so the same symbol literal always points to the same object (which is why (eq? 'leet 'leet) works). But two separate (quote leet) lists are distinct objects in memory, so eq? returns #f.

The Fix #

If you remove the inner quotes from your list literals, you'll get the behavior you expect. Use '(leet a f) instead of '('leet 'a 'f):

(eq? (first '(leet a f)) (first '(leet coder a f f))) ; returns #t

Now both first calls return the same interned symbol leet, so eq? correctly identifies them as identical.

Bonus: If You Did Want to Compare Those 'leet Values #

If for some reason you intended to compare the 'leet list objects (though that's probably not your goal here), you could use equal? instead of eq?—it checks for structural equality rather than identity:

(equal? (first '('leet 'a 'f)) (first '('leet 'coder 'a 'f 'f))) ; returns #t

内容的提问来源于stack exchange，提问作者msafadieh