How can I access a shadowed definition in Forth?

Question

When a word is redefined, is it possible to access the old word?

Imagine there is a word foo defined and redefined

: foo ( n -- 2*n ) 2* ;  ok
: foo ( n -- 2*n+1 ) foo 1+ ; redefined foo   ok
10 foo . 21  ok

foo here executed both definitions.

Is it possible to execute the first definition ("second-foo")?

21 second-foo . 42 ok

To see it?

see second-foo
: foo
  2* ; ok

Answer 1

A simple way to access a shadowed definition is to create a synonym for the definition before creating a new definition with the same name.

synonym old-foo foo
: foo ... ;

Another way is to use introspection tools, namely the word traverse-wordlist . Using this word we can define a word find-name-nth-in that is similar to the standardized word find-name-in ( c-addr u wid -- nt|0 ) , but it finds n-th shadowed word, as follows:

: find-name-nth-in ( c-addr1 u1 u wid -- nt|0 )
  >r 1+
  [: ( sd u nt -- sd u true | sd nt 0 false )
    2>r 2dup r@ name>string compare if rdrop r> true exit then
    r> r> 1- dup if nip then dup 0<>
  ;] r> traverse-wordlist ( sd u | sd nt 0 )
  if 2drop 0 exit then nip nip
;

This find-name-nth-in word returns an nt for the word that has the given name in the given word list, and after which exactly u words with the same name (case-sensitively) were defined in the word list , or 0 otherwise.

A test case:

: ?name ( nt|0 -- nt ) dup 0= -13 and throw ;

: foo ." old" ;
: foo ." new" ;

"foo" 0 forth-wordlist find-name-nth-in ?name name>interpret execute
\ prints "new"

"foo" 1 forth-wordlist find-name-nth-in ?name name>interpret execute
\ prints "old"

In Gforth the word xt-see can be used as:

"foo" 1 forth-wordlist find-name-nth-in ?name name>interpret xt-see