hyungi
e1da984e08
문제 SR과 카드 SR이 같은 간격 상수·산술을 참조하도록 순수함수 추출. 운영 동작 무변경.
- app/services/study/sr_schedule.py: REVIEW_INTERVAL_DAYS{1:3,2:7,3:14}/MASTERED=4/FIRST_DUE=1
+ advance(stage,outcome,now)→(new_stage,new_due) | None(skipped) + first_due(now).
진입 게이트(due_at IS NOT NULL/최초 due/skipped 불변)는 호출부 잔류(finalize vs review-complete 정책 차이).
- session_finalize.py: 상수·advance 분기 → sr_schedule import + sr_advance() (re-export 유지).
- study_question_progress.py: DEFAULT_FIRST_DUE_DAYS → sr_schedule import.
- 회귀 테스트 7/7: 전진 1·3·7·14·졸업·리셋·skipped불변·상수 + 전 stage×outcome 구 로직 바이트 동등.
Co-Authored-By: Claude Opus 4.8 (1M context) <noreply@anthropic.com>
94 lines
3.3 KiB
Python
94 lines
3.3 KiB
Python
"""sr_schedule 공용추출 회귀 테스트 (B1).
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문제 SR 동작이 추출 전과 동일함을 보장 — advance() 가 구 session_finalize 분기 로직과
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바이트 동등(전진 1·3·7·14 / 졸업 / 오답 리셋 / skipped 불변 / 상수값)인지 검증.
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"""
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from __future__ import annotations
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import sys
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from datetime import datetime, timedelta, timezone
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from pathlib import Path
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ROOT = Path(__file__).resolve().parent.parent
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sys.path.insert(0, str(ROOT / "app"))
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from services.study import sr_schedule as sr # noqa: E402
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NOW = datetime(2026, 6, 7, 12, 0, 0, tzinfo=timezone.utc)
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def _old_logic(review_stage, outcome, now):
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"""추출 전 session_finalize.py:188-201 의 산술을 그대로 재현 (동등성 기준)."""
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if outcome == "correct":
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new_stage = (review_stage or 0) + 1
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if new_stage >= 4:
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return new_stage, None
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return new_stage, now + timedelta(days={1: 3, 2: 7, 3: 14}[new_stage])
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elif outcome in ("wrong", "unsure"):
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return 0, now + timedelta(days=1)
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return None # skipped
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def test_constants():
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assert sr.REVIEW_INTERVAL_DAYS == {1: 3, 2: 7, 3: 14}
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assert sr.REVIEW_STAGE_MASTERED == 4
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assert sr.DEFAULT_FIRST_DUE_DAYS == 1
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def test_advance_correct_progression():
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assert sr.advance(None, "correct", NOW) == (1, NOW + timedelta(days=3))
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assert sr.advance(0, "correct", NOW) == (1, NOW + timedelta(days=3))
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assert sr.advance(1, "correct", NOW) == (2, NOW + timedelta(days=7))
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assert sr.advance(2, "correct", NOW) == (3, NOW + timedelta(days=14))
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def test_advance_graduation():
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# stage 3 → correct → stage 4 = 졸업(due_at=None)
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assert sr.advance(3, "correct", NOW) == (4, None)
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assert sr.advance(4, "correct", NOW) == (5, None)
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def test_advance_reset():
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assert sr.advance(0, "wrong", NOW) == (0, NOW + timedelta(days=1))
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assert sr.advance(2, "wrong", NOW) == (0, NOW + timedelta(days=1))
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assert sr.advance(2, "unsure", NOW) == (0, NOW + timedelta(days=1))
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def test_advance_skipped_no_change():
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assert sr.advance(1, "skipped", NOW) is None
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assert sr.advance(3, "skipped", NOW) is None
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def test_first_due():
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assert sr.first_due(NOW) == (0, NOW + timedelta(days=1))
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def test_equivalence_with_old_logic():
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# 전 stage × 전 outcome 조합에서 추출 함수 == 구 로직.
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for stage in (None, 0, 1, 2, 3, 4):
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for outcome in ("correct", "wrong", "unsure", "skipped"):
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assert sr.advance(stage, outcome, NOW) == _old_logic(stage, outcome, NOW), \
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f"mismatch stage={stage} outcome={outcome}"
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def test_reexport_preserved():
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# 기존 import 경로 (session_finalize / study_question_progress) 가 상수를 재-export.
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from services.study import session_finalize as sf
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assert sf.REVIEW_INTERVAL_DAYS == sr.REVIEW_INTERVAL_DAYS
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assert sf.REVIEW_STAGE_MASTERED == sr.REVIEW_STAGE_MASTERED
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assert sf.DEFAULT_FIRST_DUE_DAYS == sr.DEFAULT_FIRST_DUE_DAYS
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_TESTS = [v for k, v in dict(globals()).items() if k.startswith("test_")]
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if __name__ == "__main__":
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# session_finalize import 는 무거운 의존(ai 등) 가능 — reexport 테스트만 조건부.
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ran = 0
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for t in _TESTS:
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if t.__name__ == "test_reexport_preserved":
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continue
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t()
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ran += 1
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print(f"OK ({ran} pure tests; reexport는 pytest에서)")
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