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\begin{document}
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\noindent{\Large\bfseries Cel karty}\par
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Uczeń przechodzi od liniowego kursora z K\&R 5.4 do wielokrotnego użycia tej samej areny: rozpoznaje nagłówek bloku, first-fit, split oraz dwustronne scalanie wolnych bloków w heap\_4.
\vspace{0.8em}
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\vspace{0.7em}
\noindent{\Large\bfseries Zakres karty}\par
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Dwa małe modele C przygotowują przewidywanie. Główny replay uruchamia prawdziwe FreeRTOS-Kernel V11.3.0 heap\_4 na RV32I/Hazard3 i dowodzi fragmentacji, pełnego scalenia, minimum-ever oraz kontrolowanego OOM.
\vspace{0.55em}
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\textbf{Lekcja} & \textbf{Task} & \textbf{Najważniejsza idea} & \textbf{Priorytet} & \textbf{Status} & \textbf{Version} \\ \hline
\hline
\textbf{\texttt{FC01}} & \textbf{\texttt{Task01}} & \textbf{headers, address-ordered free list, first-fit, split, coalescing and heap statistics} & \textbf{główny} & \textbf{opracowane} & \textbf{\texttt{v00.01}} \\ \hline
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\normalsize
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\noindent{\small\bfseries A1 — Od liniowego kursora do odzyskiwalnej areny \hfill część 1/1 \textperiodcentered\ r1 c1}\par
\vspace{0.35em}
\begin{figure}[H]
\centering
\includegraphics[width=0.47121\linewidth,trim=0.000bp 0.000bp 0.000bp 0.000bp,clip]{\detokenize{../assets/a1-context.png}}
\caption{A1 CONTEXT — od żądania aplikacji do rzeczywistych bajtów ucHeap. — część 1/1 (r1 c1).}
\label{fig:a1-context}
\end{figure}
\Needspace{6\baselineskip}
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Aplikacja korzysta z publicznego API, heap\_4 zarządza jedną areną, a Hazard3 pokazuje jej rzeczywisty stan.
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\noindent{\small\bfseries A2 — Nagłówek, payload i lista wolnych bloków \hfill część 1/1 \textperiodcentered\ r1 c1}\par
\vspace{0.35em}
\begin{figure}[H]
\centering
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\caption{A2 STRUCTURE — metadane wewnątrz areny i adresowo uporządkowana lista. — część 1/1 (r1 c1).}
\label{fig:a2-structure}
\end{figure}
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Każdy blok ma nagłówek, a tylko wolne bloki wykorzystują wskaźnik next do budowy listy uporządkowanej adresami.
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\noindent{\small\bfseries A5 — First-fit, fragmentacja i pełne scalenie \hfill część 1/1 \textperiodcentered\ r1 c1}\par
\vspace{0.35em}
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\centering
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\caption{A5 FLOW — inicjalizacja, A/B/C, fragmentacja, dwustronne scalenie i OOM. — część 1/1 (r1 c1).}
\label{fig:a5-flow}
\end{figure}
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Jeden przebieg A/B/C przechodzi od pustej areny przez trzy przydziały i dwa wolne obszary do ponownie scalonego bloku.
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\noindent{\small\bfseries A6 — Stan bloku a stan statystyk \hfill część 1/1 \textperiodcentered\ r1 c1}\par
\vspace{0.35em}
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\centering
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\caption{A6 STATE — wolny, przydzielony, rozdzielony, scalony oraz minimum-ever. — część 1/1 (r1 c1).}
\label{fig:a6-state}
\end{figure}
\Needspace{6\baselineskip}
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Bieżąca liczba wolnych bajtów może rosnąć po free, ale minimum-ever pozostaje historycznym minimum.
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\noindent{\small\bfseries A7 — Dowód w Hazard3 i ELF \hfill część 1/1 \textperiodcentered\ r1 c1}\par
\vspace{0.35em}
\begin{figure}[H]
\centering
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\caption{A7 RUNTIME — kod, ELF, checkpoint, pamięć i statystyki jednego przebiegu. — część 1/1 (r1 c1).}
\label{fig:a7-runtime}
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Ten sam artefakt łączy źródło, symbole ELF, osiem checkpointów, pamięć ucHeap i końcowe asercje.
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\section{Task01 — mechanika heap\_4}
\begin{ESCTaskFrame}{TASK01 · Predict and prove reusable heap\_4 blocks}
{\scriptsize\ttfamily aea09ab7-9bfd-5368-8d1b-c8724871c1a7}\par
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Przejdź przez model nagłówka, split i porządek adresowy przed uruchomieniem prawdziwego heap\_4.
\begin{ESCStepFrame}{size · Policz rozmiary całych bloków.}
Rozróżnij surowy BlockLink\_t (8 bajtów) od xHeapStructSize (16 bajtów), dodaj nagłówek do żądań 24, 40 i 16 bajtów, a następnie wyrównaj wynik do 16.
\end{ESCStepFrame}
\begin{ESCStepFrame}{list · Narysuj listę po free(A,C).}
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\end{ESCBlockFrame}
\begin{ESCBlockFrame}{B — replay Hazard3}
Powiąż każdy stan diagramu z deterministycznym checkpointem E01--E08.
\begin{ESCStepFrame}{allocate · Sprawdź E01--E04.}
Potwierdź wartości 4080, 4032, 3968 i 3936 oraz zużycie 48/64/32 bajtów.
\end{ESCStepFrame}
\begin{ESCStepFrame}{release · Sprawdź E05--E07.}
Porównaj liczbę wolnych bloków i udowodnij scalenie poprzedniego, a potem następnego sąsiada.
\end{ESCStepFrame}
\begin{ESCStepFrame}{verify · Sprawdź E08.}
Udowodnij zachowanie minimum-ever, kontrolowany OOM, brak asercji i PASS=1.
\end{ESCStepFrame}
\end{ESCBlockFrame}
\begin{ESCBlockFrame}{Ćwiczenie · Ćwiczenie — ta sama suma, inna użyteczność}
Zaproponuj dwa układy wolnych bloków o tej samej sumie bajtów, ale tylko jeden zdolny obsłużyć wskazane duże żądanie. Uzasadnij odpowiedź przez largest-free-block.
\par\textbf{Evidence:} Dwa diagramy listy, suma wolnych bajtów, rozmiar największego bloku i wynik pvPortMalloc.
\par\textbf{Acceptance:} Uczeń nie utożsamia sumy wolnej pamięci z możliwością pojedynczego dużego przydziału.
\end{ESCBlockFrame}
\tcblower\textbf{Task acceptance:} Na RV32 BlockLink\_t ma 8 bajtów, lecz 16-bajtowe wyrównanie portu daje xHeapStructSize=16; A/B/C zużywają 48/64/32 bajty; po free(A,C) są dwa wolne bloki, po free(B) jeden; final free = initial free; minimum-ever nie rośnie; OOM zwraca NULL; PASS=1.
\end{ESCTaskFrame}
\begin{ESCConclusionFrame}
heap\_4 odzyskuje pamięć dzięki metadanym w arenie, adresowo uporządkowanej liście i scalaniu sąsiadów. Suma wolnych bajtów nie zastępuje informacji o największym bloku, a minimum-ever jest historią, nie bieżącym stanem.
\end{ESCConclusionFrame}
\ESCSectionBlockEnd
\end{document}
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