flan/test/programs/dev-rerun.flan

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;;;; What a re-run does to a global, which is decided by the form that
;;;; defined it and not by the daemon.
;;;;
;;;; A [defvar] is Common Lisp's [defvar]: its initialiser runs only if the
;;;; variable is not already initialised, so its value survives a re-run. A
;;;; plain zeroed one always did — .bss is untouched by a second entry into
;;;; main — and a computed one did not, because the startup function [main]
;;;; calls ran again from the top and stored the initial value back over
;;;; whatever the last run had left. A [defconst] is a constant and the
;;;; question does not arise.
;;;;
;;;; So each line printed below is a claim about one of those cases, and the
;;;; run number is the first of them: [runs] is computed, so before the fix it
;;;; counted 1, 1, 1.
(import agent "vendor:agent")
(defconst base i64 40)
;; Computed, and the whole reproduction: the initialiser is a call, so it is
;; lifted into the startup function rather than written into the image.
(defn start [] i64 base)
(defvar counter i64 (start))
;; Zero-valued, which needs no startup at all and must keep needing none.
(defvar zeroed i64)
;; A computed dyn global: the map is built by a function, rooted before the
;; startup function runs, and mutated by every run. Its contents have to
;; survive a re-run for the same reason [counter]'s value does, and its root
;; has to survive collection either way.
(defn table [] dyn {:runs 0})
(defvar state dyn (table))
(defn main [] i32
(agent/start "/tmp/flan-dev-rerun-fallback.sock")
(set counter (+ counter 1))
(set zeroed (+ zeroed 2))
(put state :runs (+ (get state :runs) 1))
(print "counter ") (print counter) (println "")
(print "zeroed ") (print zeroed) (println "")
(print "runs ") (print (get state :runs)) (println "")
(print "base ") (print base) (println "")
;; Long enough for a client to be served, short enough to park well inside
;; any watchdog — dev-macro.flan's clock, for its reason.
(dotimes [i 200]
(agent/wait 5))
0)