flan/test/programs/strings.flan
Joseph Ferano 02850d7282 The prelude returns new bytes now: a builder, join, split and the case pair
Eight functions that the file used to refuse by name, and the refusal was
always one sentence -- there is no allocator -- which stopped being true when
Vec landed. Three rules hold across all of them and are written at the head
of the section: the result is owned and the caller frees it, the allocator is
the context's, and no signature carries a Result because no allocating
operation returns an error.

The builder is not a type. Odin's strings.Builder wraps a [dynamic]u8; here
the (Vec u8) already is that and already has push, so a wrapper would be a
move-only struct whose only method is the one it wraps. What was missing is
appending a run of bytes, and append! is that -- taking a (Ptr (Vec u8)),
because a Vec parameter moves and a by-value builder would be consumed by its
first append.

append-i64! and append-f64! are the argument for the whole shape. The
runtime renders numbers into one shared static buffer, so two of its results
cannot be held at once; these copy out before returning, so a builder holds as
many numbers as it likes. strings.flan puts two integers and a float on one
line to show it.

split returns a (Vec [u8]) and not a (Vec (Vec u8)): the fields borrow the
input, and the owning shape is refused outright because a Vec copies and
releases its elements bytewise. Constructing it needed a one-line slices-new,
because (vec-new) takes its element type as a bare symbol and [u8] is not
one -- a compiler gap, noted rather than worked around in silence.

replace-bytes guards its empty needle with an if and not an early return: a
returned Vec is a move, the dead set spans the function, and a return on one
branch would kill the binding on the other.
2026-09-12 21:51:42 +07:00

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;;;; The prelude's second tier: the functions that return new storage.
;;;;
;;;; Every one of these was refused by name in prelude.ml until there was an
;;;; allocator to return a Vec from, and this file is the corpus that says the
;;;; refusals are lifted. The cases are chosen the way the slice-algorithm
;;;; tests were: each is an input a plausible wrong version gets wrong.
;;;;
;;;; Everything allocated here is freed, even though leaking is defined
;;;; behaviour (spec-memory.md), because this file is the example people copy.
;;; A (Vec u8) printed as text, without the caller writing the two-step every
;;; time. as-slice borrows -- it copies ptr+len and never the elements -- so v
;;; is still the owner afterwards and is still free-able.
(defn show [v (Ptr (Vec u8))]
(println (string (as-slice (deref v)))))
(defn main [] i32
;; The builder. Three appends and two numbers into one Vec, which is the
;; case the shared static scratch buffer in the runtime makes impossible for
;; i64->bytes on its own: two of its results cannot be held at once, and
;; these two numbers are both in the answer.
(let [b (vec-new u8)]
(append! (addr b) (bytes "x="))
(append-i64! (addr b) 42)
(append! (addr b) (bytes " y="))
(append-i64! (addr b) -7)
(append! (addr b) (bytes " r="))
(append-f64! (addr b) 1.5)
(show (addr b)) ; x=42 y=-7 r=1.5
(free b))
;; concat over three parts, and over none -- the empty result rather than a
;; trap.
(let [parts [(bytes "one") (bytes "") (bytes "two")]]
(let [c (concat (slice parts 0 3))]
(show (addr c)) ; onetwo
(free c)))
(let [parts [(bytes "unused")]]
(let [c (concat (slice parts 0 0))]
(println (len c)) ; 0
(free c)))
;; join: n parts, n-1 separators. The one-part case is the one that must not
;; emit a separator at all, and the zero-part case is the one a "append then
;; chop the tail" join gets wrong because there is no tail.
(let [parts [(bytes "a") (bytes "b") (bytes "c")]]
(let [j (join (slice parts 0 3) (bytes ", "))]
(show (addr j)) ; a, b, c
(free j))
(let [j (join (slice parts 0 1) (bytes ", "))]
(show (addr j)) ; a
(free j))
(let [j (join (slice parts 0 0) (bytes ", "))]
(println (len j)) ; 0
(free j))
;; An empty separator is concat.
(let [j (join (slice parts 0 3) (bytes ""))]
(show (addr j)) ; abc
(free j)))
;; repeat, including zero times.
(let [r (repeat-bytes (bytes "ab") 3)]
(show (addr r)) ; ababab
(free r))
(let [r (repeat-bytes (bytes "ab") 0)]
(println (len r)) ; 0
(free r))
;; The allocating case pair. The input is a string literal, which lives in
;; .rodata -- an in-place lower would either segfault at -O0 or be deleted at
;; -O2, and that is exactly why these exist. Digits and punctuation pass
;; through untouched, which is the range check a table-free version gets
;; wrong by shifting every byte.
(let [l (to-lower (bytes "Hello, World 42!"))]
(show (addr l)) ; hello, world 42!
(free l))
(let [u (to-upper (bytes "Hello, World 42!"))]
(show (addr u)) ; HELLO, WORLD 42!
(free u))
;; replace. "aaa" with "aa" -> "b" is the non-overlapping rule: the answer is
;; "ba", because the match consumes both a's and the scan resumes after them.
(let [r (replace-bytes (bytes "aaa") (bytes "aa") (bytes "b"))]
(show (addr r)) ; ba
(free r))
;; A replacement longer than what it replaces, and one that is empty.
(let [r (replace-bytes (bytes "a,b,c") (bytes ",") (bytes " -- "))]
(show (addr r)) ; a -- b -- c
(free r))
(let [r (replace-bytes (bytes "a,b,c") (bytes ",") (bytes ""))]
(show (addr r)) ; abc
(free r))
;; No occurrence is a copy, and an empty `from` is a copy -- the reading
;; where it matches everywhere is an infinite loop.
(let [r (replace-bytes (bytes "abc") (bytes "z") (bytes "!"))]
(show (addr r)) ; abc
(free r))
(let [r (replace-bytes (bytes "abc") (bytes "") (bytes "!"))]
(show (addr r)) ; abc
(free r))
;; split. n separators, n+1 fields, always -- so the trailing empty field is
;; present, which is where Odin's own iterator and its allocating split
;; disagree with each other.
(let [f (split (bytes "a,b,c") \,)]
(println (len f)) ; 3
(println (string (at f 0))) ; a
(println (string (at f 2))) ; c
(free f))
(let [f (split (bytes "a,b,") \,)]
(println (len f)) ; 3
(println (len (at f 2))) ; 0
(free f))
(let [f (split (bytes ",a") \,)]
(println (len f)) ; 2
(println (len (at f 0))) ; 0
(free f))
;; No separator at all is one field, and the empty input is one empty field.
(let [f (split (bytes "abc") \,)]
(println (len f)) ; 1
(println (string (at f 0))) ; abc
(free f))
(let [f (split (bytes "") \,)]
(println (len f)) ; 1
(println (len (at f 0))) ; 0
(free f))
;; The fields are slices of the input and nothing was copied: this one
;; round-trips through join, and the separator it rebuilds with is a
;; different one, so an implementation that handed back the original slice
;; would print the original string.
(let [f (split (bytes "a,b,c") \,)]
(let [j (join (as-slice f) (bytes "/"))]
(show (addr j)) ; a/b/c
(free j))
(free f))
;; The allocator is the context's, so with-allocator moves the whole tier
;; into an arena -- which is the answer to the fixed arity of a defn, and the
;; reason none of these takes an allocator argument. free-all releases every
;; one of them at once, including the Vec still bound below it.
(let [a (arena-new 4096)]
(with-allocator a
(let [parts [(bytes "in") (bytes "arena")]]
(let [j (join (slice parts 0 2) (bytes "-"))]
(show (addr j)) ; in-arena
;; Not freed: an arena cannot release one block, and free-all is
;; what releases this.
(free j))))
(free-all a)
(arena-destroy a))
0)