functional programming - Destructive operations in scheme environments -

i'm confused destructive operations in scheme. let's have list , destructive procedures defined in global environment:

(define '(a b c)) (define (mutate-obj x)     (set! x '(mutated))) (define (mutate-car! x)     (set-car! x 'mutated)) (define (mutate-cdr! x)     (set-cdr! x 'mutated)) 

then have following expression evaulation:

(mutate-obj! a) => (a b c) (mutate-car! a) => (mutated b c) (mutate-cdr! a) => (mutated . mutated) 

why isn't set! having effect on a outside procedure when both set-car! , set-cdr! have? why isn't expression on first line evaluating (mutated)? how of work?

the first example isn't working imagined. although both x parameter , a global variable pointing same list, when execute (set! x '(mutated)) set x (a parameter local procedure) point different list, , a remains unchanged. it'd different if wrote this:

(define (mutate-obj)   (set! '(mutated))) 

now a gets mutated inside procedure. second , third procedures modifying contents of a list, pointed x, change gets reflected "outside" once procedure returns.