Attribute Grammars

Given CFG $G = (N,T,P,S)$ , Attribute grammar is $(G,S,I,R)$

and are synthesized and inherited attributes of $X \in N$ resp.

$\begin{displaymath}A(X) = S(X) \cup I(X) \end{displaymath}$
$r \in R(p)$ are ``rules'' for each $p \in P = X_0 : X_1 \ldots X_n$ .

$\begin{displaymath}X_{i_0}.a_0 = f(X_{i_1}.a_1,\ldots,X_{i_k}) \end{displaymath}$
Can only define inherited attributes of right-hand side symbols and synthesized attributes of left-hand side symbol:

$\begin{displaymath}(i_0 = 0 \wedge a_0 \in S(X_{i_0})) \vee (i_0 > 0 \wedge a_0 \in I(X_{i_0})) \end{displaymath}$

Meaning of Attribute Grammars

Given a tree over the context-free grammar, assume all nodes have a unique name.
Instantiate wherever $p \in P$ occurs in the tree, replacing $X_{i_j}$ with the name of the corresponding node.
Solve resulting equations simultaneously.

APS syntax (part 1)

Terminals are ``types'':
$\begin{program} type $T$\ = \ldots; \end{program}$
Nonterminals are ``phyla'':
$\begin{program} phylum $P$; \end{program}$
Productions are ``constructors'':
$\begin{program} constructor $c$($f_1$:$T_1$,\ldots,$f_n$:$T_n$); \end{program}$

APS syntax (part 2)

Attributes have declared types and defaults. Inherited attributes must be so declared:
$\begin{program} attribute $P$.$a$\ : $T$\ := $e$; pragma inherited(a); \end{program}$
Rules are ``pattern matches'':
$\begin{program} match ?$v_0$=$c$(?$v_1$,\ldots,?$v_n$) begin \ldots $v_{i_0}$.a := $e$; \ldots end; \end{program}$
The expression may use attribute values, constants etc.

Tiny example

$\begin{program} \par module TINY[] begin phylum Wood; \par constructor branch(x... ...b=branch(?x,?y) begin b.leaves := x.leaves + y.leaves; end; end; \end{program}$

John Tang Boyland
2004-09-30