我对Common Lisp中的函数副本有疑问.
在Scheme中我会选择:
(define (foo par1 par2) (+ par1 par2))
(define bar foo)
(print (bar 1 2)) ;; --> prints 3
(define (foo par1 par2) (* par1 par2))
(print (bar 1 2)) ;; --> prints again 3
(print (foo 1 2)) ;; --> prints 2
我怎么能用Common Lisp做到这一点？