Introduce normal type placeholders \ntv

This commit is contained in:
Andreas Stadelmeier 2024-02-12 15:28:44 +01:00
parent 5074c21943
commit fca93d7ec6
2 changed files with 42 additions and 29 deletions

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@ -111,6 +111,7 @@
\newcommand{\wtypestore}[3]{\ensuremath{#1 = \wtype{#2}{#3}}}
%\newcommand{\wtype}[2]{\ensuremath{[#1\ #2]}}
\newcommand{\wtv}[1]{\ensuremath{\tv{#1}_?}}
\newcommand{\ntv}[1]{\ensuremath{\underline{\tv{#1}}}}
\newcommand{\wcstore}{\ensuremath{\Delta}}
%\newcommand{\rwildcard}[1]{\ensuremath{\mathtt{#1}_?}}

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@ -1,6 +1,18 @@
\section{Unify}\label{sec:unify}
The wildcard placeholders are used for intermediat types.
It is not possible to create all super types of a type.
The General rule only creates the ones expressable by Java syntax, which still are infinitly many in some cases \cite{TamingWildcards}.
%thats not true. it can spawn X^T_T2.List<X> where T and T2 are types and we need to choose one inbetween them
Otherwise the algorithm could generate more solutions, but they have to be filterd out afterwards, because they cannot be translated into Java.
Any type can be inserted into wildcard placeholders.
Normal placeholders have to contain types, which are well-formed under the supplied environment.
% It is important that the algorithm also works for any subset of constraints
%TODO: The unify algorithm can do any operation on wildcard placeholders the same as on normal ones.
%TODO: Unify could make way more substitutions for wtvs especially in step 2
\subsection{Description}
The \unify{} algorithm tries to find a solution for a set of constraints like
$\set{\exptype{List}{String} \lessdot \tv{a}, \exptype{List}{Integer} \lessdot \tv{a}}$.
@ -137,13 +149,13 @@ This is necessary for the \rulename{Equals} rule to work properly.
\begin{tabular}[t]{l@{~}l}
\rulename{Subst} &
$\begin{array}[c]{l}
\wildcardEnv \vdash C \cup \set{\tv{a} \doteq \type{T}}\\
\wildcardEnv \vdash C \cup \set{\ntv{a} \doteq \type{T}}\\
\hline
[\type{T}/\tv{a}]\wildcardEnv \vdash [\type{T}/\tv{a}]
C \cup \set{\tv{a} \doteq \type{T}}
C \cup \set{\ntv{a} \doteq \type{T}}
\end{array}
\quad \begin{array}{c}
\tv{a} \notin \type{T} \\
\ntv{a} \notin \type{T} \\
\text{fv}(\type{T}) \subseteq \Delta'
\end{array}$\\
\\
@ -184,7 +196,7 @@ $
\wildcardEnv \cup \set{\wildcard{A}{U}{L}} \vdash C \cup \, \set{ \type{U} \lessdot \type{G} }
\end{array}
\quad \quad
\begin{array}[c]{l}
\begin{array}[c]{l} %TODO: can the second part be removed by adding a X.C<X> <. C<a?> constraint at method invocation?
\wildcardEnv \cup \set{\wildcard{A}{U}{L}} \vdash C \cup \, \set{ \type{A} \lessdotCC \type{G} } \\
\hline
\vspace*{-0.4cm}\\
@ -329,24 +341,24 @@ $
The \rulename{match} rule generates fresh wildcards $\overline{\wildcard{A}{\tv{u}}{\tv{l}}}$.
Their upper and lower bounds are fresh type variables.
%Unify only renames the wildcards in the reduce rule
% It's the only place where wildcards are coming into play (theres always a reduce step before a wildcard substitution is possible)
% die wildcard variablen sollten erst am Ende ausgetauscht werden gegen normale variablen
% das funktioniert, da die im Reduce step erstellten direkt substituiert werden
% die anderen erlauben Capture Conversion aber nur wenn der Methodentyp und Parametertyp schon feststeht! (gleich Mächtig wie TI in Java)
% a? <. T ->
% T <. a? ->
% a? =. T -> substitute!
% bei normalen Typvariablen werden keine Wildcards substituiert
% \begin{tcolorbox}
% $
% \wctype{\rwildcard{X}}{Box}{\rwildcard{X}} \lessdot \exptype{Box}{\tv{a}_?}, \\
% \exptype{Box}{\tv{a}_?} \lessdot \wctype{\rwildcard{X}}{Box}{\rwildcard{X}}
% $
% \end{tcolorbox}
\begin{figure}
\begin{center}
\leavevmode
\fbox{
\begin{tabular}[t]{l@{~}l}
\rulename{Circle} & $
\begin{array}[c]{l}
\wildcardEnv \vdash C \cup \, \set{\tv{a}_1 \lessdot
\tv{a}_2, \tv{a}_2 \lessdot \tv{a}_3, \dots, \tv{a}_n \lessdot \tv{a}_1}\\
\hline
\wildcardEnv \vdash C \cup \, \set{\tv{a}_1 \doteq \tv{a}_2, \tv{a}_2 \doteq \tv{a}_3, \dots , \tv{a}_n \doteq \tv{a}_1}
\end{array} \quad n>0
$
\end{tabular}}
\end{center}
\caption{Rules for normal placeholders}\label{fig:reduce-rules}
\end{figure}
\begin{figure}
\begin{center}
@ -751,10 +763,10 @@ This builds a search tree over multiple possible solutions.
& $
\begin{array}[c]{l}
\wildcardEnv \vdash C \cup \set{ \tv{a} \lessdot \type{N},
\tv{a} \lessdot^* \tv{b}}
\tv{a} \lessdot \tv{b}}
\\
\hline
\wildcardEnv \vdash C \cup \set{ \tv{a} \lessdot^* \tv{b}, \tv{b} \lessdot \type{N} }
\wildcardEnv \vdash C \cup \set{ \tv{a} \lessdot \tv{b}, \tv{b} \lessdot \type{N} }
\end{array}
$
\\\\
@ -1049,13 +1061,13 @@ Otherwise the generation rules \rulename{GenSigma} and \rulename{GenDelta} will
\rulename{GenDelta}
& $
\deduction{
\wildcardEnv \vdash C \cup \set{\tv{b} \lessdot \type{T} } \implies \Delta, \sigma
\wildcardEnv \vdash C \cup \set{\ntv{b} \lessdot \type{T} } \implies \Delta, \sigma
}{
\wildcardEnv \vdash [\type{B}/\tv{b}]C \implies \Delta \cup \set{\wildcard{B}{\type{T}}{\bot}}, \sigma \cup \set{\tv{b} \to \type{B}}
\wildcardEnv \vdash [\type{B}/\ntv{b}]C \implies \Delta \cup \set{\wildcard{B}{\type{T}}{\bot}}, \sigma \cup \set{\ntv{b} \to \type{B}}
} \quad
\begin{array}{l}
\tph(\type{T}) = \emptyset, \text{fv}(\type{T}) \subseteq \Delta \cup \Delta_{in} \\
\rwildcard{B} \ \text{fresh}, \tv{b} \notin \text{dom}(\sigma), \Delta, \Delta_{in} \vdash \type{T} \ \ok
\rwildcard{B} \ \text{fresh}, \ntv{b} \notin \text{dom}(\sigma), \Delta, \Delta_{in} \vdash \type{T} \ \ok
\end{array}
$
\\\\
@ -1063,14 +1075,14 @@ Otherwise the generation rules \rulename{GenSigma} and \rulename{GenDelta} will
& $
\deduction{
\wildcardEnv \vdash C \cup
\set{\tv{a} \doteq \type{T} } \implies \Delta, \sigma
\set{\ntv{a} \doteq \type{T} } \implies \Delta, \sigma
}{
\wildcardEnv \vdash C \implies \Delta, \sigma \cup
\set{\tv{a} \to \type{T} }
\set{\ntv{a} \to \type{T} }
} \quad
\begin{array}{l}
\tph(\type{T}) = \emptyset \\ %,\, \text{fv}(\type{T}) \subseteq \Delta \\ % T ok implies that
\tv{a} \notin \text{dom}(\sigma),\, \Delta, \Delta_{in} \vdash \type{T} \ \ok
\ntv{a} \notin \text{dom}(\sigma),\, \Delta, \Delta_{in} \vdash \type{T} \ \ok
\end{array}
$
\\\\