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Solution gen methods. Do not generate Generics

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Andreas Stadelmeier 2024-06-18 18:45:29 +02:00
parent 0a1644617a
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73
aspUnify.tex
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@ -94,29 +94,52 @@ We want to bring type inference for Java to the next level.
} }
\end{mathpar} \end{mathpar}
\begin{mathpar}
\inferrule[N-Refl]{}{
\type{C} << \type{C}
}
\and
\inferrule[N-Trans]{\type{T}_1 << \type{T}_2 \\ \type{T}_2 << \type{T}_3}{
\type{T}_1 << \type{T}_3
}
\and
\inferrule[N-Class]{\texttt{class}\ \exptype{C}{\ldots} \triangleleft \exptype{D}{\ldots}}{
\type{C} << \type{D}
}
\end{mathpar}
\section{Unify} \section{Unify}
Input: Every type placeholder must have an upper bound. Input: Every type placeholder must have an upper bound.
Output: Every $\tv{a} \lessdot \type{T}$ constraint gets a Output: Every $\tv{a} \lessdot \type{T}$ constraint gets a
The $\tv{a} \lessdot \type{Object}$ rule has to be ensured by the incoming constraints.
The need to have an upper bound to every type placeholder.
We have to formulate the unification algorithm with implication rules. We have to formulate the unification algorithm with implication rules.
Those can be directly translated to ASP. Those can be directly translated to ASP.
\begin{mathpar} \begin{mathpar}
\inferrule[Subst]{
\tv{a} \doteq \type{N}
}{
\tv{a} \mapsto \type{N}
}
\and
\inferrule[Subst-L]{ \inferrule[Subst-L]{
\tv{a} \doteq \type{T}_1 \\ \tv{a} \mapsto \type{T}_1 \\
\tv{a} \lessdot \type{T}_2 \tv{a} \lessdot \type{T}_2
}{ }{
\type{T}_1 \lessdot \type{T}_2 \type{T}_1 \lessdot \type{T}_2
} }
\and \and
\inferrule[Subst-R]{ \inferrule[Subst-R]{
\tv{a} \doteq \type{T}_1 \\ \tv{a} \mapsto \type{T}_1 \\
\type{T}_2 \lessdot \tv{a} \type{T}_2 \lessdot \tv{a}
}{ }{
\type{T}_2 \lessdot \type{T}_1 \type{T}_2 \lessdot \type{T}_1
} }
\and \and
\inferrule[Subst-Equal]{ \inferrule[Subst-Equal]{
\tv{a} \doteq \type{T}_1 \\ \tv{a} \mapsto \type{T}_1 \\
\tv{a} \doteq \type{T}_2 \tv{a} \doteq \type{T}_2
}{ }{
\type{T}_1 \doteq \type{T}_2 \type{T}_1 \doteq \type{T}_2
@ -135,10 +158,10 @@ Those can be directly translated to ASP.
} }
\and \and
\inferrule[Subst-Param]{ \inferrule[Subst-Param]{
\tv{a} \doteq \type{G} \\ \tv{a} \mapsto \type{S} \\
\type{T} \doteq \exptype{C}{\type{T}_1 \ldots, \tv{a}, \ldots \type{T}_n} \\ \type{T} \doteq \exptype{C}{\type{T}_1 \ldots, \tv{a}, \ldots \type{T}_n} \\
}{ }{
\type{T} \doteq \exptype{C}{\type{T}_1, \ldots \type{G}, \ldots \type{T}_n} \type{T} \doteq \exptype{C}{\type{T}_1, \ldots \type{S}, \ldots \type{T}_n}
} }
\and \and
\inferrule[S-Object]{}{\tv{a} \lessdot \type{Object}} \inferrule[S-Object]{}{\tv{a} \lessdot \type{Object}}
@ -225,20 +248,35 @@ Result:
\sigma(\tv{a}) = \type{N} \sigma(\tv{a}) = \type{N}
} }
\and \and
\inferrule[Solution-Sub]{ \inferrule[Generic]{
\tv{a} \lessdot \exptype{C_1}{\ol{T_1}}, \ldots, \tv{a} \lessdot \exptype{C_n}{\ol{T_n}} \\ \tv{a} \lessdot \type{N} %, \ldots, \tv{a} \lessdot \exptype{C_n}{\ol{T_n}} \\
\forall i: \type{C_m} << \type{C_i} \\ \\
\text{not}\ \tv{a} \doteq \type{N} \text{not}\ \tv{a} \doteq \type{N}'
}{ }{
\Delta(\type{A}) = \exptype{C_m}{\ol{T_m}} \\ \tv{a} \mapsto \type{A} % \Delta(\type{A}) = \exptype{C_m}{\ol{T_m}} \\
\tv{a} \mapsto \type{A}
}
% \and
% \inferrule[Solution-Sub]{
% \tv{a} \lessdot \exptype{C_1}{\ol{T_1}}, \ldots, \tv{a} \lessdot \exptype{C_n}{\ol{T_n}} \\
% \forall i: \type{C_m} << \type{C_i} \\
% \text{not}\ \tv{a} \doteq \type{N}
% }{
% \Delta(\type{A}) = \exptype{C_m}{\ol{T_m}} \\ \sigma(\tv{a}) = \type{A}
% }
\and
\inferrule[Solution-Gen]{
\tv{a} \lessdot \type{G}_1, \ldots, \tv{a} \lessdot \type{G}_n \\
\forall i: \type{G} <: \type{G}_i \\
}{
\Delta(\type{A}) = \type{G} \\ \sigma(\tv{a}) = \type{A}
} }
\and \and
\inferrule[Solution-Sub]{ \inferrule[Solution-Gen]{
\tv{a} \lessdot \exptype{C_1}{\ol{T_1}}, \ldots, \tv{a} \lessdot \exptype{C_n}{\ol{T_n}} \\ \tv{a} \lessdot \type{C}_1, \ldots, \tv{a} \lessdot \type{C}_n \\
\forall i: \type{C_m} << \type{C_i} \\ \forall i: \type{C}_m << \type{C}_i \\
\text{not}\ \tv{a} \doteq \type{N}
}{ }{
\Delta(\type{A}) = \exptype{C_m}{\ol{T_m}} \\ \sigma(\tv{a}) = \type{A} \tv{a} \doteq \type{C}_m
} }
\end{mathpar} \end{mathpar}
@ -288,6 +326,11 @@ Fail:
The algorithm terminates if every type placeholder in the input constraint set has an assigned type. The algorithm terminates if every type placeholder in the input constraint set has an assigned type.
\section{ASP Encoding}
\begin{tabular}{l | r}
$\exptype{C}{\ol{X}}$ & \texttt{type("C", paramX)}\\
\end{tabular}
\section{Completeness} \section{Completeness}
To proof completeness we have to show that every type can be replaced by a placeholder in a correct constraint set. To proof completeness we have to show that every type can be replaced by a placeholder in a correct constraint set.