Joseph Ferano
a0f37e72a2
The !-means-mutates convention distinguished nothing — there is no immutable counterpart to contrast with — so every mutating name drops the mark: sort, sort-by, sort-bytes, swap, reverse, append, append-i64, append-f64, encode-rune, split-next, map-remove, map-next, and the test helpers beside them. Two could not simply shed it: map! is map-in-place, because map is the into transform's word and means the non-mutating thing; put! is put-at, because put is the Map builtin. The ?-means-asks convention stays. Dated records keep the old spellings; watch.clj's reset-spies! and the other Clojure names are not ours to rename.
155 lines
6.5 KiB
Plaintext
155 lines
6.5 KiB
Plaintext
;;;; The prelude's second tier: the functions that return new storage.
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;;;;
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;;;; Every one of these was refused by name in prelude.ml until there was an
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;;;; allocator to return a Vec from, and this file is the corpus that says the
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;;;; refusals are lifted. The cases are chosen the way the slice-algorithm
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;;;; tests were: each is an input a plausible wrong version gets wrong.
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;;;;
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;;;; Everything allocated here is freed, even though leaking is defined
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;;;; behaviour (spec-memory.md), because this file is the example people copy.
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;;; A (Vec u8) printed as text, without the caller writing the two-step every
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;;; time. as-slice borrows -- it copies ptr+len and never the elements -- so v
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;;; is still the owner afterwards and is still free-able.
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(defn show [v (Ptr (Vec u8))] ()
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(println (string (as-slice (deref v)))))
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(defn main [] i32
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;; The builder. Three appends and two numbers into one Vec, which is the
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;; case the shared static scratch buffer in the runtime makes impossible for
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;; i64->bytes on its own: two of its results cannot be held at once, and
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;; these two numbers are both in the answer.
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(let [b (vec-new u8)]
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(append (addr b) (bytes "x="))
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(append-i64 (addr b) 42)
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(append (addr b) (bytes " y="))
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(append-i64 (addr b) -7)
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(append (addr b) (bytes " r="))
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(append-f64 (addr b) 1.5)
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(show (addr b)) ; x=42 y=-7 r=1.5
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(free b))
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;; concat over three parts, and over none -- the empty result rather than a
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;; trap.
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(let [parts [(bytes "one") (bytes "") (bytes "two")]]
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(let [c (concat (slice parts 0 3))]
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(show (addr c)) ; onetwo
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(free c)))
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(let [parts [(bytes "unused")]]
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(let [c (concat (slice parts 0 0))]
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(println (len c)) ; 0
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(free c)))
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;; join: n parts, n-1 separators. The one-part case is the one that must not
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;; emit a separator at all, and the zero-part case is the one a "append then
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;; chop the tail" join gets wrong because there is no tail.
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(let [parts [(bytes "a") (bytes "b") (bytes "c")]]
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(let [j (join (slice parts 0 3) (bytes ", "))]
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(show (addr j)) ; a, b, c
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(free j))
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(let [j (join (slice parts 0 1) (bytes ", "))]
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(show (addr j)) ; a
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(free j))
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(let [j (join (slice parts 0 0) (bytes ", "))]
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(println (len j)) ; 0
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(free j))
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;; An empty separator is concat.
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(let [j (join (slice parts 0 3) (bytes ""))]
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(show (addr j)) ; abc
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(free j)))
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;; repeat, including zero times.
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(let [r (repeat-bytes (bytes "ab") 3)]
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(show (addr r)) ; ababab
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(free r))
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(let [r (repeat-bytes (bytes "ab") 0)]
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(println (len r)) ; 0
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(free r))
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;; The allocating case pair. The input is a string literal, which lives in
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;; .rodata -- an in-place lower would either segfault at -O0 or be deleted at
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;; -O2, and that is exactly why these exist. Digits and punctuation pass
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;; through untouched, which is the range check a table-free version gets
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;; wrong by shifting every byte.
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(let [l (to-lower (bytes "Hello, World 42!"))]
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(show (addr l)) ; hello, world 42!
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(free l))
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(let [u (to-upper (bytes "Hello, World 42!"))]
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(show (addr u)) ; HELLO, WORLD 42!
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(free u))
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;; replace. "aaa" with "aa" -> "b" is the non-overlapping rule: the answer is
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;; "ba", because the match consumes both a's and the scan resumes after them.
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(let [r (replace-bytes (bytes "aaa") (bytes "aa") (bytes "b"))]
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(show (addr r)) ; ba
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(free r))
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;; A replacement longer than what it replaces, and one that is empty.
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(let [r (replace-bytes (bytes "a,b,c") (bytes ",") (bytes " -- "))]
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(show (addr r)) ; a -- b -- c
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(free r))
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(let [r (replace-bytes (bytes "a,b,c") (bytes ",") (bytes ""))]
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(show (addr r)) ; abc
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(free r))
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;; No occurrence is a copy, and an empty `from` is a copy -- the reading
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;; where it matches everywhere is an infinite loop.
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(let [r (replace-bytes (bytes "abc") (bytes "z") (bytes "!"))]
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(show (addr r)) ; abc
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(free r))
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(let [r (replace-bytes (bytes "abc") (bytes "") (bytes "!"))]
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(show (addr r)) ; abc
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(free r))
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;; split. n separators, n+1 fields, always -- so the trailing empty field is
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;; present, which is where Odin's own iterator and its allocating split
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;; disagree with each other.
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(let [f (split (bytes "a,b,c") \,)]
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(println (len f)) ; 3
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(println (string (at f 0))) ; a
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(println (string (at f 2))) ; c
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(free f))
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(let [f (split (bytes "a,b,") \,)]
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(println (len f)) ; 3
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(println (len (at f 2))) ; 0
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(free f))
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(let [f (split (bytes ",a") \,)]
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(println (len f)) ; 2
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(println (len (at f 0))) ; 0
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(free f))
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;; No separator at all is one field, and the empty input is one empty field.
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(let [f (split (bytes "abc") \,)]
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(println (len f)) ; 1
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(println (string (at f 0))) ; abc
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(free f))
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(let [f (split (bytes "") \,)]
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(println (len f)) ; 1
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(println (len (at f 0))) ; 0
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(free f))
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;; The fields are slices of the input and nothing was copied: this one
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;; round-trips through join, and the separator it rebuilds with is a
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;; different one, so an implementation that handed back the original slice
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;; would print the original string.
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(let [f (split (bytes "a,b,c") \,)]
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(let [j (join (as-slice f) (bytes "/"))]
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(show (addr j)) ; a/b/c
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(free j))
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(free f))
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;; The allocator is the context's, so with-allocator moves the whole tier
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;; into an arena -- which is the answer to the fixed arity of a defn, and the
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;; reason none of these takes an allocator argument. free-all is what
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;; releases the region, and arena-destroy hands it back.
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(let [a (arena-new 4096)]
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(with-allocator a
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(let [parts [(bytes "in") (bytes "arena")]]
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(let [j (join (slice parts 0 2) (bytes "-"))]
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(show (addr j)) ; in-arena
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;; The free is written because the binding is dead after it either
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;; way, and it keeps the block: an arena cannot release one, which
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;; is the difference the capability set exists to state. free-all
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;; below is what actually releases this.
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(free j))))
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(free-all a)
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(arena-destroy a))
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0)
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