At Wed, 26 Apr 2017 15:04:15 -0700 (PDT), Dupéron Georges wrote:
> However, I'm not sure what operations can cause compile-time code to
> be run in this situation. My (possibly incorrect) understanding is
> that macros are executed only once (when expanding the code), but the
> code to the right-hand-side of a (define-syntax foo
> costly-operation), and the code within a (begin-for-syntax …) will
> both be run each time the module is required.

Each time a module is compiled, the compile-time code for any imported
modules is run in fresh compile-time instantiations of those modules.
("Compile" = "expand" in this context.)

So, given "a.rkt" as

 ;; a.rkt
 #lang racket
 (require "b.rkt")

and "b.rkt" as

 ;; b.rkt
 #lang racket
 (begin-for-syntax (displayln "b"))

if you use `racket a.rkt` leaving both modules as source, you'll see
printed "b" twice: once when compiling "b.rkt", and once when compiling
"a.rkt". If you use `raco make a.rkt`, you'll also see "b" printed
twice, but then `racket a.rkt` will not print "b".

If you have run `raco make a.rkt` already, then here's a potentially
confusing result:

 % racket
 Welcome to Racket
 > (require "a.rkt")
 > 1

In this case, no "b" is printed to load and execute the runtime part of
"a.rkt". Still, compile-time module instances have been created to
handle further evaluation in the REPL. So, as soon as `eval` is called
by the REPL for the expression `1`, the compile-time parts of "a.rkt"
and "b.rkt" are run. That's why "b" prints, and why it prints so late.

When you've compiled a program to bytecode and then run it, whether any
compile-time code is run depends on whether you use `eval` or
`dynamic-require` or similar.

Another part of the run time is just reading in bytecode, though. The
bytecode-loading part of Racket seems slower than it should be, and
that was one of the things I had planned to rebuild --- but now I'm
hoping that we'll get better load times by swapping Chez Scheme in
place of the current runtime system.

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