Start Lecture #8

Remark: midterm grades are based on lab1.

4.6.5: Viable Prefixes


4.7: More Powerful LR Parsers

We consider very briefly two alternatives to SLR, canonical-LR or LR, and lookahead-LR or LALR.

4.7.1: Canonical LR(1) Items

SLR used the LR(0) items, that is the items used were productions with an embedded dot, but contained no other (lookahead) information. The LR(1) items contain the same productions with embedded dots, but add a second component, which is a terminal (or $). This second component becomes important only when the dot is at the extreme right (indicating that a reduction can be made if the input symbol is in the appropriate FOLLOW set). For LR(1) we do that reduction only if the input symbol is exactly the second component of the item. This finer control of when to perform reductions, enables the parsing of a larger class of languages.

4.7.2: Constructing LR(1) Sets of Items


4.7.3: Canonical LR(1) Parsing Tables


4.7.4: Constructing LALR Parsing Tables

For LALR we merge various LR(1) item sets together, obtaining nearly the LR(0) item sets we used in SLR. LR(1) items have two components, the first, called the core, is a production with a dot; the second a terminal. For LALR we merge all the item sets that have the same cores by combining the 2nd components (thus permitting reductions when any of these terminals is the next input symbol). Thus we obtain the same number of states (item sets) as in SLR since only the cores distinguish item sets.

Unlike SLR, we limit reductions to occurring only for certain specified input symbols. LR(1) gives finer control; it is possible for the LALR merger to have reduce-reduce conflicts when the LR(1) items on which it is based is conflict free.

Although these conflicts are possible, they are rare and the size reduction from LR(1) to LALR is quite large. LALR is the current method of choice for bottom-up, shift-reduce parsing.

4.7.5: Efficient Construction of LALR Parsing Tables


4.7.6: Compaction of LR Parsing Tables


4.8: Using Ambiguous Grammars


4.8.1: Precedence and Associativity to Resolve Conflicts


4.8.2: The Dangling-Else Ambiguity


4.8.3: Error Recovery in LR Parsing


4.9: Parser Generators

4.9.1: The Parser Generator Yacc

The tool corresponding to Lex for parsing is yacc, which (at least originally) stood for yet another compiler compiler. This name is cute but somewhat misleading since yacc (like the previous compiler compilers) does not produce a compiler, just a parser.

The structure of the yacc user input is similar to that for lex, but instead of regular definitions, one includes productions with semantic actions.

There are ways to specify associativity and precedence of operators. It is not done with multiple grammar symbols as in a pure parser, but more like declarations.

Use of Yacc requires a serious session with its manual.

4.9.2: Using Yacc with Ambiguous Grammars


Creating Yacc Lexical Analyzers with Lex


4.9.4: Error Recovery in Yacc


Chapter 5: Syntax-Directed Translation

Homework: Read Chapter 5.

Again we are redoing, more formally and completely, things we briefly discussed when breezing over chapter 2.

Recall that a syntax-directed definition (SDD) adds semantic rules to the productions of a grammar. For example to the production T → T1 / F we might add the rule
  T.code = T1.code || F.code || '/'
if we were doing an infix to postfix translator.

Rather than constantly copying ever larger strings to finally output at the root of the tree after a depth first traversal, we can perform the output incrementally by embedding semantic actions within the productions themselves. The above example becomes
  T → T1 / F { print '/' }

Since we are generating postfix, the action comes at the end (after we have generated the subtrees for T1 and F, and hence performed their actions). In general the actions occur within the production, not necessarily after the last symbol.

For SDD's we conceptually need to have the entire tree available after the parse so that we can run the postorder traversal. (It is postorder since we have a simple S-attributed SDD; we will traverse the parse tree in other orders when the SDD is not S-attributed, and will see situations when no traversal order is possible.)

In some situations semantic actions can be performed during the parse, without saving the tree.

5.1: Syntax-Directed Definitions (SDDs)

Formally, attributes are values (of any type) that are associated with grammar symbols. Write X.a for the attribute a of symbol X. You can think of attributes as fields in a record/struct/object.

Semantic rules (rules for short) are associated with productions.

5.1.1: Inherited and Synthesized Attributes

Terminals can have synthesized attributes; these are given to it by the lexer (not the parser). For example the token id might well have the attribute lexeme, where id.lexeme is the lexeme that the lexer converted into this instance of id. There are no rules in an SDD giving attribute values to terminals. Terminals do not have inherited attributes (to be defined shortly).

A nonterminal B can have both inherited and synthesized attributes. The difference is how they are computed. In either case the computation is specified by a rule associated with the production at a node N of the parse tree.

Definition: A synthesized attribute of a nonterminal B, is defined at a node N (where B is the LHS of the production associated with N). The attribute can depend on only (synthesized or inherited) attribute values at the children of N (the RHS of the production) and on inherited attribute values at N itself.

The arithmetic division example above was synthesized.

ProductionSemantic Rules

L → E $L.val = E.val
E → E1 + TE.val = E1.val + T.val
E → E1 - TE.val = E1.val - T.val
E → TE.val = T.val
T → T1 * FT.val = T1.val * F.val
T → T1 / FT.val = T1.val / F.val
T → FT.val = F.val
F → ( E )F.val = E.val
F → numF.val = num.lexval

Example: The SDD at the right gives a left-recursive grammar for expressions with an extra nonterminal L added as the start symbol. The terminal num is given a value by the lexer, which corresponds to the value stored in the numbers table for lab 2.

Draw the parse tree for 7+6/3 on the board and verify that L.val is 9, the value of the expression.

This example uses only synthesized attributes.

Definition: An SDD with only synthesized attributes is called S-attributed and has the property that the rules give the attribute of the LHS in terms of attributes of the RHS.

For an S-attributed SDD, all the rules can be evaluated by a single bottom-up (i.e., postorder) traversal of the annotated parse tree.

Inherited attributes are more complicated since the node N of the parse tree with which the attribute is associated (and which is also the natural node to store the value) does not contain the production with the corresponding semantic rule.

Definition: An inherited attribute of a nonterminal B at node N (where B is the LHS of the production) is defined by a semantic rule of the production at the parent of N (where B occurs in the RHS of the production). The value depends only on attributes at N, N's siblings, and N's parent.

Note that when viewed from the parent node P (the site of the semantic rule), the inherited attribute depends on values at P and at P's children (the same as for synthesized attributes). However, and this is crucial, the nonterminal B is the LHS of a child of P and hence the attribute is naturally associated with that child. It is possibly stored there and is shown there in the diagrams below.

We will see examples with inherited attributes soon.

Definition: Often the attributes are just evaluations without side effects. In such cases we call the SDD an attribute grammar.

bad sdd

5.1.2: Evaluating an SDD at the Nodes of a Parse Tree

If we are given an SDD and a parse tree for a given sentence, we would like to evaluate the annotations at every node. Since, for synthesized annotations parents can depend on children, and for inherited annotations children can depend on parents, there is no guarantee that one can in fact find an order of evaluation. The simplest counterexample is the single production A→B with synthesized attribute A.syn, inherited attribute B.inh, and rules A.syn=B.inh and B.inh=A.syn+1. This means to evaluate A.syn at the parent node we need B.inh at the child and vice versa. Even worse it is very hard to tell, in general, if every sentence has a successful evaluation order.

All this not withstanding we will not have great difficulty because we will not be considering the general case.


Annotated Parse Trees

Recall that a parse tree has leaves that are terminals and internal nodes that are non-terminals.

Definition: A parse tree decorated with attributes, is called an annotated parse tree It is constructed as follows.

Each internal node corresponds to a production with the symbol labeling the node the LHS of the production. If there are no attributes for the LHS in this production, we leave the node as it was (I don't believe this is a common occurrence). If there are k attributes for the LHS, we replace the LHS in the parse tree by k equations. The LHS of the equation is the attribute and the right hand side is its value. Note that the annotated parse tree contains all the information of the original parse tree since we replaced something like E with something like E.att=7.

We computed the values to put in this tree for 7+6/3 and on the right is (7-6).

Homework: 1

3*5*4 left rec

Why Have Inherited Attributes?

Consider the following left-recursive grammar for multiplication of numbers and the parse tree on the right for 3*5*4.

    T → T * F
    T → F
    F → num

It is easy to see how the values can be propagated up the tree and the expression evaluated.

When doing top-down parsing, we need to avoid left recursion. Consider the grammar below, which is the result of removing the left recursion, and again its parse tree is shown on the right. Try not to look at the semantic rules for the moment. 3*5*4
ProductionSemantic RulesType

T → F T'T'.lval = F.valInherited
T.val = T'.tvalSynthesized

T' → * F T1' T'1.lval = T'.lval * F.valInherited
T'.tval = T'1.tvalSynthesized

T' → εT'.tval = T'.lvalSynthesized

F → numF.val = num.lexvalSynthesized

Where on the tree should we do the multiplication 3*5? There is no node that has 3 and * and 5 as children. The second production is the one with the * so that is the natural candidate for the multiplication site. Make sure you see that this production (for the * in 3*5) is associated with the blue-highlighted node in the parse tree. The right operand (5) can be obtained from the F that is the middle child of this T'. F gets the value from its child, the number itself; this is an example of the simple synthesized case we have already seen, F.val=num.lexval (see the last semantic rule in the table).

But where is the left operand? It is located at the sibling of T' in the parse tree, i.e., at the F immediately to T's left (we shall see the significance that the sibling is to the left; it is not significant that it is immediately to the left). This F is not mentioned in the production associated with the T' node we are examining. So, how does T' get F.val from its sibling? The common parent, in this case T, can get the value from F and then our node can inherit the value from its parent.
Bingo! ... an inherited attribute. This can be accomplished by having the following two rules at the node T.
    T.tmp   = F.val (synthesized)
    T'.lval = T.tmp (inherited)

Since we have no other use for T.tmp, we combine the above two rules into the first rule in the table.

Now lets look at the second multiplication (3*5)*4, where the parent of T' is another T'. (This is the normal case. When there are n multiplies, n-1 have T' as parent and only one has T).

The pink-highlighted T' is the site for the multiplication. However, it needs as left operand, the product 3*5 that its parent can calculate. So we have the parent (another T' node, the blue one in this case) calculate the product and store it as an attribute of its right child namely the pink T'. That is the first rule for T' in the table.

We have now explained the first, third, and last semantic rules. These are enough to calculate the answer. Indeed, if we trace it through, 60 does get evaluated and stored in the bottom right T', the one associated with the ε-production. Our remaining goal is to get the value up to the root where it represents the evaluation of this term T and can be combined with other terms to get the value of a larger expression.

3*5*4 annotated

Going up is easy, just synthesize. I named the attribute tval, for term-value. It is generated at the ε-production from the lval attribute (which at this node is not a good name) and propagated back up. At the T node it is called simply val. At the right we see the annotated parse tree for this input.

Homework: Extend this SDD to handle the left-recursive, more complete expression evaluator given earlier in this section. Don't forget to eliminate the left recursion first.

It clearly requires some care to write the annotations.

Another question is how does the system figure out the evaluation order, assuming one exists? That is the subject of the next section.

Remark: Consider the identifier table. The lexer creates it initially, but as the compiler performs semantic analysis and discovers more information about various identifiers, e.g., type and visibility information, the table is updated. One could think of this as some inherited/synthesized attribute pair that during each phase of analysis is pushed down and back up the tree. However, it is not implemented this way; the table is made a global data structure that is simply updated. The compiler writer must ensure manually that the updates are performed in an order respecting any dependences.

5.2: Evaluation Orders for SDD's

5.2.1: Dependency Graphs


The diagram on the right illustrates a great deal. The black dotted lines comprise the parse tree for the multiplication grammar just studied when applied to a single multiplication, e.g. 3*5. Each synthesized attribute is shown in green and is written to the right of the grammar symbol at the node where it is defined. Each inherited attribute is shown in red and is written to the left of the grammar symbol where it is defined.

Each green arrow points to the synthesized attribute calculated from the attribute at the tail of the arrow. These arrows either go up the tree one level or stay at a node. That is because a synthesized attribute can depend only on the node where it is defined and that node's children. The computation of the attribute is associated with the production at the node at its arrowhead. In this example, each synthesized attribute depends on only one other, but that is not required.

Each red arrow points to the inherited attribute calculated from the attribute at the tail. Note that two red arrows point to the same attribute. This indicates that the common attribute at the arrowheads, depends on both attributes at the tails. According to the rules for inherited attributes, these arrows either go down the tree one level, go from a node to a sibling, or stay within a node. The computation of the attribute is associated with the production at the parent of the node at the arrowhead.

5.2.2: Ordering the Evaluation of Attributes

The graph just drawn is called the dependency graph. In addition to being generally useful in recording the relations between attributes, it shows the evaluation order(s) that can be used. Since the attribute at the head of an arrow depends on the on the one at the tail, we must evaluate the head attribute after evaluating the tail attribute.

Thus what we need is to find an evaluation order respecting the arrows. This is called a topological sort. The rule is that the needed ordering can be found if and only if there are no (directed) cycles. The algorithm is simple.

  1. Choose a node having no incoming edges
  2. Delete the node and all outgoing edges.
  3. Repeat
If the algorithm terminates with nodes remaining, there is a directed cycle within these remaining nodes and hence no suitable evaluation order.

If the algorithm succeeds in deleting all the nodes, then the deletion order is a suitable evaluation order and there were no directed cycles.

The topological sort algorithm is nondeterministic (Choose a node) and hence there can be many topological sort orders.

Homework: 1.

5.2.3: S-Attributed Definitions

Given an SDD and a parse tree, it is easy to tell (by doing a topological sort) whether a suitable evaluation exists (and to find one).

However, a very difficult problem is, given an SDD, are there any parse trees with cycles in their dependency graphs, i.e., are there suitable evaluation orders for all parse trees. Fortunately, there are classes of SDDs for which a suitable evaluation order is guaranteed.

As mentioned above an SDD is S-attributed if every attribute is synthesized. For these SDDs all attributes are calculated from attribute values at the children since the other possibility, the tail attribute is at the same node, is impossible since the tail attribute must be inherited for such arrows. Thus no cycles are possible and the attributes can be evaluated by a postorder traversal of the parse tree.

Since postorder corresponds to the actions of an LR parser when reducing the body of a production to its head, it is often convenient to evaluate synthesized attributes during an LR parse.


5.2.4 L-Attributed Definitions

Unfortunately, it is hard to live without inherited attributes. So we often restrict the inherited attributes to a subclass for which we can easily find an evaluation order. Fortunately, this class is sufficient for our needs.

Definition: An SDD is L-Attributed if each attribute is either
  1. Synthesized.
  2. Inherited from the left, and hence the name L-attributed.
    Specifically, if the production is A → X1X2...Xn, then the inherited attributes for Xj can depend only on
    1. Inherited attributes of A, the LHS.
    2. Any attribute of X1, ..., Xj-1, i.e. only on symbols to the left of Xj.
    3. Attributes of Xj, *BUT* you must guarantee (separately) that the attributes of Xj do not by themselves cause a cycle.

Case 2c must be handled specially whenever it occurs. The top picture to the right illustrates the other cases and suggests why there cannot be any cycles.

The picture immediately to the right corresponds to a fictitious R-attributed definition. One reason L-attributed definitions are favored over R, is the left to right ordering in English. See the example below on type declarations and also consider the grammars that result from eliminating left recursion.

We shall see that the key property of L-attributed SDDs is that they can be evaluated with two passes over the tree (an euler-tour order) in which we evaluate the inherited attributes as we go down the tree and the synthesized attributes as we go up. The restrictions L-attributed SDDs place on the inherited attributes are just enough to guarantee that when we go down we have all the values needed for the inherited attributes of the child.

ProductionSemantic Rule

A → B C B.inh = A.inh
C.ihn = A.inh - B.inh + B.syn
A.syn = A.inh * B.inh + B.syn - C.inh / C.syn

B → XX.inh = something
B.syn = B.inh + X.syn

C → YY.inh = something
C.syn = C.inh + Y.syn

Evaluating L-Attributed Definitions

The table on the right shows a very simple grammar with fairly general, L-attributed semantic rules attached. Compare the dependencies with the general case shown in the (red-green) picture of L-attributed SDDs above.

The picture below the table shows the parse tree for the grammar in the table. The triangles below B and C represent the parse tree for X and Y. The dotted and numbered arrow in the picture illustrates the evaluation order for the attributes; it will be discussed shortly.
eval order l-attr

The rules for calculating A.syn, B.inh, and C.inh are shown in the table. The attribute A.inh would have been set by the parent of A in the tree; the semantic rule generating A.inh would be given with the production at the parent. The attributes X.syn and Y.syn are calculated at the children of B and C respectively. X.syn can depend of X.inh and on values in the triangle below X; similarly for Y.syn.

The picture to the right shows that there is an evaluation order for L-attributed definitions (again assuming no case 2c). We just need to follow the (Euler-tour) arrow and stop at all the numbered points. As in the pictures above, red signifies inherited attributes and green synthetic. Specifically, the evaluations at the numbered stops are

  1. A is invoked (viewing the traversal as a program) and is passed its inherited attributes (A.inh in our case, but of course there could be several such attributes), which have been evaluated at its parent.
  2. B is invoked by A and is given B.inh, which A has calculated. In programming terms: A executes
        call B(B.inh)
    where the argument has been evaluated by A. This argument can depend on A.inh since the parent of A has given A this value.
  3. B calls its first child (in our example X is the only child) and passes to the child its inherited attributes. In programming terms: B executes
        call X(X.inh)
  4. The child returns passing back to B. the synthesized attributes of the child. In programming terms: X executes
        return X.syn
    In reality there could be more synthesized attributes, there could be more children, the children could have children, etc.
  5. B returns to A passing back B.syn, which can depend on B.inh (given to B by A in step 2) and X.syn (given to B by X in the previous step).
  6. A calls C giving C its inherited attributes, which can depend on A.ihn (given to A, by A's parent), B.inh (previously calculated by A in step 2), and B.syn (given to A by B in step 5).
  7. C calls its first child, just as B did.
  8. The child returns to C, just as B's child returned to B.
  9. C returns to A passing back C.syn, just as B did.
  10. A returns to its parent passing back A.syn, which can depend on A.inh (given to A by its parent in step 1), B.inh calculated by A in step 2, B.syn (given to A by B in step 5), C.inh (calculated by A in step 6), and C.syn (given to A by C in step 9).

More formally, do a depth first traversal of the tree and evaluate inherited attributes on the way down and synthetic attributes on the way up. This corresponds to a an Euler-tour traversal. It also corresponds to a call graph of a program where actions are taken at each call and each return

The first time you visit a node (on the way down), evaluate its inherited attributes (in programming terms, the parent evaluates the inherited attributes of the children and passes them as arguments to the call). The second time you visit a node (on the way back up), you evaluate the synthesized attributes (in programming terms the child returns the value to the parent).

eval l-attr

The key point is that all attributes needed will have already been evaluated. Consider the rightmost child of the root in the diagram on the right.

  1. Inherited attributes (which are evaluated on the first, i.e., downward, pass): An inherited attribute depends only on inherited attributes from the parent and on (inherited or synthesized) attributes from left siblings.
  2. Synthesized attributes (which are evaluated on the second, i.e., upward pass): A synthesized attribute depends only on (inherited or synthesized) attributes of its children and on its own inherited attributes.

Homework: 3(a-c).

5.2.5: Semantic Rules with Controlled Side Effects

ProductionSemantic RuleType

D → T LL.type = T.typeinherited

T → INTT.type = integersynthesized

T → FLOATT.type = floatsynthesized

L → L1 , ID L1.type = L.typeinherited
addType(ID.entry,L.type)synthesized, side effect

L → IDaddType(ID.entry,L.type)synthesized, side effect

When we have side effects such as printing or adding an entry to a table we must ensure that we have not added a constraint to the evaluation order that causes a cycle.

For example, the left-recursive SDD shown in the table on the right propagates type information from a declaration to entries in an identifier table.

The function addType adds the type information in the second argument to the identifier table entry specified in the first argument. Note that the side effect, adding the type info to the table, does not affect the evaluation order.

Draw the dependency graph on the board. Note that the terminal ID has an attribute (given by the lexer) entry that gives its entry in the identifier table. The nonterminal L has (in addition to L.type) a dummy synthesized attribute, say AddType, that is a place holder for the addType() routine. AddType depends on the arguments of addType(). Since the first argument is from a child, and the second is an inherited attribute of this node, we have legal dependences for a synthesized attribute.

Thus we have an L-attributed definition.

Homework: For the SDD above, give the annotated parse tree for

    INT a,b,c

5.3: Applications of Syntax-Directed Translations

5.3.1: Construction of Syntax Trees

ProductionSemantic Rules

E → E 1 + T E.node = new Node('+',E1.node,T.node)
E → E 1 - T E.node = new Node('-',E1.node,T.node)
E → TE.node = T.node
T → ( E )T.node = E.node
T → IDT.node = new Leaf(ID,ID.entry)
T → NUMT.node = new Leaf(NUM,NUM.val)

Recall that in syntax tree (technically an abstract syntax tree) has just the essentials. For example 7+3*5, would have one + node, one *, and the three numbers. Lets see how to construct the syntax tree from an SDD.

Assume we have two functions Leaf(op,val) and Node(op,c1,...,cn), that create leaves and interior nodes respectively of the syntax tree. Leaf is called for terminals. Op is the label of the node (op for operation) and val is the lexical value of the token. Node is called for nonterminals and the ci's refer (are pointers) to the children.

ProductionSemantic RulesType

E → T E'E.node=E'.synSynthesized

E' → + T E'1 E'1.node=new Node('+',E'.node,T.node)Inherited

E' → - T E'1 E'1.node=new Node('-',E'.node,T.node)Inherited

E' → εE'.syn=E'.nodeSynthesized
T → ( E )T.node=E.nodeSynthesized
T → IDT.node=new Leaf(ID,ID.entry)Synthesized
T → NUMT.node=new Leaf(NUM,NUM.val)Synthesized

The upper table on the right shows a left-recursive grammar that is S-attributed (so all attributes are synthesized).

Try this for x-2+y and see that we get the syntax tree.