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(*<*)
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theory CodeGen = Main:
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(*>*)
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section{*Case Study: Compiling Expressions*}
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text{*\label{sec:ExprCompiler}
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The task is to develop a compiler from a generic type of expressions (built
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from variables, constants and binary operations) to a stack machine. This
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generic type of expressions is a generalization of the boolean expressions in
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\S\ref{sec:boolex}. This time we do not commit ourselves to a particular
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type of variables or values but make them type parameters. Neither is there
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a fixed set of binary operations: instead the expression contains the
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appropriate function itself.
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*}
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types 'v binop = "'v \<Rightarrow> 'v \<Rightarrow> 'v";
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datatype ('a,'v)expr = Cex 'v
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| Vex 'a
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| Bex "'v binop" "('a,'v)expr" "('a,'v)expr";
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text{*\noindent
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The three constructors represent constants, variables and the application of
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a binary operation to two subexpressions.
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The value of an expression with respect to an environment that maps variables to
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values is easily defined:
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*}
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consts value :: "('a,'v)expr \<Rightarrow> ('a \<Rightarrow> 'v) \<Rightarrow> 'v";
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primrec
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"value (Cex v) env = v"
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"value (Vex a) env = env a"
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"value (Bex f e1 e2) env = f (value e1 env) (value e2 env)";
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text{*
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The stack machine has three instructions: load a constant value onto the
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stack, load the contents of an address onto the stack, and apply a
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binary operation to the two topmost elements of the stack, replacing them by
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the result. As for @{text"expr"}, addresses and values are type parameters:
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*}
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datatype ('a,'v) instr = Const 'v
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| Load 'a
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| Apply "'v binop";
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text{*
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The execution of the stack machine is modelled by a function
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@{text"exec"} that takes a list of instructions, a store (modelled as a
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function from addresses to values, just like the environment for
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evaluating expressions), and a stack (modelled as a list) of values,
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and returns the stack at the end of the execution---the store remains
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unchanged:
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*}
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consts exec :: "('a,'v)instr list \<Rightarrow> ('a\<Rightarrow>'v) \<Rightarrow> 'v list \<Rightarrow> 'v list";
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primrec
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"exec [] s vs = vs"
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"exec (i#is) s vs = (case i of
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Const v \<Rightarrow> exec is s (v#vs)
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| Load a \<Rightarrow> exec is s ((s a)#vs)
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| Apply f \<Rightarrow> exec is s ((f (hd vs) (hd(tl vs)))#(tl(tl vs))))";
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text{*\noindent
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Recall that @{term"hd"} and @{term"tl"}
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return the first element and the remainder of a list.
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Because all functions are total, @{term"hd"} is defined even for the empty
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list, although we do not know what the result is. Thus our model of the
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machine always terminates properly, although the definition above does not
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tell us much about the result in situations where @{term"Apply"} was executed
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with fewer than two elements on the stack.
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The compiler is a function from expressions to a list of instructions. Its
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definition is obvious:
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*}
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consts comp :: "('a,'v)expr \<Rightarrow> ('a,'v)instr list";
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primrec
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"comp (Cex v) = [Const v]"
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"comp (Vex a) = [Load a]"
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"comp (Bex f e1 e2) = (comp e2) @ (comp e1) @ [Apply f]";
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text{*
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Now we have to prove the correctness of the compiler, i.e.\ that the
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execution of a compiled expression results in the value of the expression:
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*}
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theorem "exec (comp e) s [] = [value e s]";
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(*<*)oops;(*>*)
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text{*\noindent
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This theorem needs to be generalized to
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*}
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theorem "\<forall>vs. exec (comp e) s vs = (value e s) # vs";
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txt{*\noindent
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which is proved by induction on @{term"e"} followed by simplification, once
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we have the following lemma about executing the concatenation of two
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instruction sequences:
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*}
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(*<*)oops;(*>*)
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lemma exec_app[simp]:
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"\<forall>vs. exec (xs@ys) s vs = exec ys s (exec xs s vs)";
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txt{*\noindent
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This requires induction on @{term"xs"} and ordinary simplification for the
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base cases. In the induction step, simplification leaves us with a formula
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that contains two @{text"case"}-expressions over instructions. Thus we add
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automatic case splitting as well, which finishes the proof:
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*}
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apply(induct_tac xs, simp, simp split: instr.split);
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(*<*)done(*>*)
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text{*\noindent
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Note that because \isaindex{auto} performs simplification, it can
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also be modified in the same way @{text simp} can. Thus the proof can be
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rewritten as
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*}
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(*<*)
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declare exec_app[simp del];
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lemma [simp]: "\<forall>vs. exec (xs@ys) s vs = exec ys s (exec xs s vs)";
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(*>*)
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apply(induct_tac xs, auto split: instr.split);
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(*<*)done(*>*)
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text{*\noindent
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Although this is more compact, it is less clear for the reader of the proof.
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We could now go back and prove \isa{exec (comp e) s [] = [value e s]}
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merely by simplification with the generalized version we just proved.
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However, this is unnecessary because the generalized version fully subsumes
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its instance.
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*}
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(*<*)
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theorem "\<forall>vs. exec (comp e) s vs = (value e s) # vs";
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by(induct_tac e, auto);
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end
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(*>*)
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