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Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
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 ! | ||