LARA

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sav07_lecture_22 [2007/06/14 22:49]
simon.blanchoud
sav07_lecture_22 [2007/06/14 22:50]
simon.blanchoud
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 Where $S_{use} = \lbrace x, y \rbrace$ as they are used in $c$ and $S_{df} = z$ as it's the newly defined variable in $c$. Where $S_{use} = \lbrace x, y \rbrace$ as they are used in $c$ and $S_{df} = z$ as it's the newly defined variable in $c$.
 Here we do not have a problem with the monotonicity of the \ operator as $S_{df}$ is not a variable. Here we do not have a problem with the monotonicity of the \ operator as $S_{df}$ is not a variable.
 +
  
  
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 Which leads to the following solutions : Which leads to the following solutions :
-\begin{displaymath}\begin{array}+\begin{displaymath}\begin{array}[l]
 & &
 [[x]] = \lbrace \lambda_y \rbrace & [[x]] = \lbrace \lambda_y \rbrace &
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 [[S_1]] & otherwise \\ [[S_1]] & otherwise \\
 \end{array} \right \} \subseteq S_2 \end{displaymath} \end{array} \right \} \subseteq S_2 \end{displaymath}
 +
  
 ===== Shape analysis of algebraic data types ===== ===== Shape analysis of algebraic data types =====
 +
 +Not seen in class, have a look at the papers !
 +
  
 ===== Using unary functions for interprocedural analysis ===== ===== Using unary functions for interprocedural analysis =====
  
-===== Solving set constraints =====+Not seen in class, have a look at the papers !
  
  
 +===== Solving set constraints =====
 +
 +Not seen in class, have a look at the papers !