Actually
compile is
not an immediate word (at least in F83). I was wrong about that. Here's the F83 definition:
: COMPILE (S -- ) R> DUP 2+ >R @ , ; \ COMPILE Compile the following word when this def. executes
This takes its return address (which points to the following word in the colon definition that called it),
dups it, adds 2 to it, and puts that back on the return stack as its new return address. Then, it fetches from its original return address with
@ (thus getting the address of the word that followed it in the colon definition, such as
?branch in my
if example above) and compiles it with
, into whatever is currently being compiled. Then, when it returns, having added 2 to the return address means that we don't actually execute
?branch or whatever; we've skipped over it.
So it doesn't change the state of the interpreter at all!
I think you're asking if you can use things like ?branch usefully without writing any immediate words. In some sense I think the answer is yes in F83 but no in standard Forth. I think you can put a code sequence like ?branch [ here 0 , ] into a colon definition to do what if does, and then later on say [ here swap ! ] to do what then does. I just typed this definition into F83, and it seems to work†:
: is3 3 = ?branch [ here 0 , ] ." yes" [ here swap ! ] ;
You could sort of think of
if and
then as being macros for
?branch [ here 0 , ] and
[ here swap ! ] respectively (although I'm omitting the checks they use for proper control structure nesting).
On the other hand, this is only possible because [ is an immediate word, and because ?branch is exposed, and happens to take an absolute address in the next word in the colon definition (as opposed to a byte delta or something). As it happens, exactly the same definition of is3 appears to work in GForth 0.7.3 and PFE 0.33.71, but it definitely will not work on, for example, any native-code-compiling Forth.
The standard way to invoke things like ?branch is using if, while, and so on. And you don't have to define any immediate words to do that, either.
______
† By "work" I mean it seems to behave the same as
: is3 3 = if ." yes" then ;