design: gate on THREE signals -- the alloc gate would have missed el #132
el #132's quadratic (strlen per character in str_char_code/str_slice) is pure CPU and allocates NOTHING. Measured on three controlled specimens: specimen allocs bytes time linear 2.00 -> O(n) 2.16 -> O(n) 2.05 -> O(n) accum 2.00 -> O(n) 3.99 -> O(n2) noisy compute FLAT FLAT 3.96 -> O(n2) 'compute' is #132's shape. A gate fitting only allocation count and bytes classifies it FLAT and passes -- it would not have caught the defect it was created for. The gate now fits time AND count AND bytes, failing if any exceeds its declared curve. Also: black_box is mandatory and consuming the result is NOT sufficient. The first 'compute' reported 0us at every n while returning a correct n2 -- clang closed the loop to a multiply. Only an opaque call restored the curve. Adds lang/tests/bench/fitprobe.el as the fitter's known-good/known-bad set, so the classifier is provable without depending on a real bug existing. Marks DESIGN.md 1.3 stale: test_compiler 3.58s -> 0.03s (119x).
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// fitprobe.el — controlled growth-curve specimens for validating the complexity fitter.
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//
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// Three deliberately-shaped workloads. None depends on a real defect existing,
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// which is the point: the fitter must be provable against KNOWN curves.
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//
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// linear — one allocation per item. count O(n), bytes O(n), time O(n)
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// accum — rebuilds its accumulator. count O(n), bytes O(n^2), time O(n^2)
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// compute — nested arithmetic, no alloc. count O(1), bytes O(1), time O(n^2)
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//
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// `compute` is the specimen that matters. It is the shape of el #132
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// (strlen-per-character inside str_char_code): pure CPU, zero allocation.
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// An allocation-only gate is structurally blind to it.
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//
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// No imports — uses runtime builtins directly so nothing collides.
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fn work_linear(n: Int) -> Int {
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let parts: [String] = native_list_empty()
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let i: Int = 0
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while i < n {
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let parts = native_list_append(parts, int_to_str(i))
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let i = i + 1
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}
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return native_list_len(parts)
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}
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fn work_accum(n: Int) -> Int {
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let acc: String = ""
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let i: Int = 0
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while i < n {
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let acc = acc + "x"
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let i = i + 1
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}
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return str_len(acc)
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}
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fn work_compute(n: Int) -> Int {
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// str_char_code is an opaque external call, so the C optimiser cannot
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// reduce this nest to a closed form the way it does with `total + 1`.
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// This is the exact shape of el #132: n scans over n characters, pure
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// CPU, ZERO allocation.
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let s: String = "abcdefghij"
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let total: Int = 0
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let i: Int = 0
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while i < n {
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let j: Int = 0
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while j < n {
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let total = total + str_char_code(s, 0)
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let j = j + 1
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}
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let i = i + 1
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}
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return total
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}
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fn run_one(mode: String, n: Int) {
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let c0: Int = el_alloc_count()
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let b0: Int = el_alloc_bytes()
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let t0: Int = el_now_instant()
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let r: Int = 0
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if str_eq(mode, "linear") { let r = work_linear(n) }
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if str_eq(mode, "accum") { let r = work_accum(n) }
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if str_eq(mode, "compute") { let r = work_compute(n) }
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let t1: Int = el_now_instant()
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let c1: Int = el_alloc_count()
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let b1: Int = el_alloc_bytes()
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println(mode + "\t" + int_to_str(n)
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+ "\t" + int_to_str(c1 - c0)
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+ "\t" + int_to_str(b1 - b0)
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+ "\t" + int_to_str((t1 - t0) / 1000)
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+ "\t" + int_to_str(r))
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return
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}
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fn sweep(mode: String) {
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run_one(mode, 200)
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run_one(mode, 400)
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run_one(mode, 800)
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run_one(mode, 1600)
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return
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}
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fn main() -> Int {
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println("mode\tn\tallocs\tbytes\tusec\tsink")
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sweep("linear")
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sweep("accum")
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sweep("compute")
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return 0
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}
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