View difference between Paste ID: FXEEbD2U and ipSkUdAk

 A list in lisp looks like this:
 
 List = [mptr 0]
 
 [mptr0] = | car0 | [mptr 1] |
 [mptr1] = | car1 | [mptr 2] |
 ...
 [mptr_n] = | car_n | NULL |
 
 So set-car! does this:
 
 Before:
 [mptr0] = | 7 | [mptr1] |
 
 (set-car! 5 [mptr0])
 
 After:
 [mptr0] = | 5 | [mptr1] |
 
 Note that we've changed the memory itself, not some amorphous "cell". We've actually manipulated memory. So now how does circular-list work? Well, let's get our same list:
 
 [mptr0] = | car0 | [mptr 1] |
 [mptr1] = | car1 | [mptr 2] |
 ...
 [mptr_n] = | car_n | NULL |
 
 Now xs here is [mptr0] when we get into circular-list. Then final-pair gives us back [mptr_n], and we change the cdr of it, yielding the following piece of arrangement:
 
 [mptr0] = | car0 | [mptr 1] |
 [mptr1] = | car1 | [mptr 2] |
 ...
 [mptr_n] = | car_n | [mptr0] |
 
 Now any time we look for the car of [mptr_n] in memory, we get the original memory address of [mptr0], and end up back at that piece of the list. Following pointers again sends us into a loop forever.
 
 eq? does magical things to understand this.