Review \(LL\) parsing
Shift-reduce parsing
The \(LR\) parsing algorithm
Constructing \(LR\) parsing tables
Top-down parsing expands a parse tree from the start symbol to the leaves
The leaves at any point form a string \(\beta A \gamma\)
\(\beta\) contains only terminals
The input string is \(\beta b \delta\)
The prefix \(\beta\) matches (is valid)
The next token is \(b\)
A predictive parser is described by a table
For each non-terminal \(A\) and for each token \(b\) we specify a production \(A \rightarrow \alpha\)
When trying to expand \(A\) we use \(A \rightarrow \alpha\) if \(b\) follows next
Once we have the table:
The parsing algorithm is simple and fast
No backtracking is necessary
Bottom-up parsing is more general than top-down parsing
and just as efficient
builds on ideas in top-down parsing
preferred method in practice
Also called \(LR\) parsing
\(L\) means that tokens are read left-to-right
\(R\) means that it constructs a rightmost derivation
\(LR\) parsers do not need left-factored grammars and can also handle left-recursive grammars
Consider the following grammar: \[E \rightarrow E + (E) \; \vert \; int\]
This is not \(LL(1)\)
Consider the string: \(int + (int) + (int)\)
\(LR\) parsing reduces a string to the start symbol by inverting productions
Given a string of terminals:
Identify \(\beta\) in the string such that \(A \rightarrow \beta\) is a production
Replace \(\beta\) by \(A\) in the string
Repeat steps 1 and 2 until the string is the start symbol (or all possibilities are exhausted)
Consider the following grammar: \[E \rightarrow E + (E) \; \vert \; int\]
And input string: \(int + (int) + (int)\)
Bottom-up parse:
\(int + (int) + (int)\)
\(E + (int) + (int)\)
\(E + (E) + (int)\)
\(E + (int)\)
\(E + (E)\)
\(E\)
A rightmost derivation in reverse
An \(LR\) parser traces a rightmost derivation in reverse
This has an interesting consequence
Let \(\alpha \beta \gamma\) be a step of a bottom-up parse
Assume the next reduction is by using \(A \rightarrow \beta\)
The \(\gamma\) is a string of terminals
This is because \(\alpha A \gamma \rightarrow \alpha \beta \gamma\) is a step in a rightmost derivation
Idea: split a string into two substrings
the right substring is the partition that has not been examined yet
the left substring has terminals and non-terminals
The dividing point is marked by a \(\vert\)
Initially, all input is unexamined: \(\vert x_1, x_2 \ldots x_n\)
Bottom-up parsing uses only two kinds of actions: shift and reduce
Shift: move \(\vert\) one place to the right \[E + (\vert int ) \rightarrow E + (int \vert)\]
Reduce: apply an inverse production at the right end of the left string
Consider the grammar: \(E \rightarrow E + (E) \; \vert \; int\)
| String | Action |
| \(\vert int + (int) + (int)\$\) | shift |
| \(int \vert + (int) + (int)\$\) | reduce \(E \rightarrow int\) |
| \(E \vert + (int) + (int)\$\) | shift three times |
| \(E + (int \vert) + (int)\$\) | reduce \(E \rightarrow int\) |
| \(E + (E \vert) + (int)\$\) | shift |
| \(E + (E) \vert + (int)\$\) | reduce \(E \rightarrow E + (E)\) |
| \(E \vert + (int)\$\) | shift three times |
| \(E + (int \vert)\$\) | reduce \(E \rightarrow int\) |
| \(E + (E \vert)\$\) | shift |
| \(E + (E) \vert\$\) | reduce \(E \rightarrow E + (E)\) |
| \(E \vert\$\) | accept |
The left string can be implemented by a stack
Shift pushes a terminal on the stack
Reduce pops zero or more symbols off of the stack (production right hand side) and pushes a non-terminal on the stack (production left hand side).
Idea: use a finite automaton (DFA) to decide when to shift or reduce
The input is the stack
The language consists of terminals and non-terminals
We run the DFA on the stack and examine the resulting state \(X\) and token \(t\) after \(\vert\)
If \(X\) has a transition labeled \(t\) then shift
If \(X\) is labeled with “\(A \rightarrow \beta\) on \(t\)” then reduce
Transitions:
\(0 \rightarrow 1\) on \(int\)
\(0 \rightarrow 2\) on \(E\)
\(2 \rightarrow 3\) on \(+\)
\(3 \rightarrow 4\) on \((\)
\(4 \rightarrow 5\) on \(int\)
\(4 \rightarrow 6\) on \(E\)
\(6 \rightarrow 7\) on \()\)
\(6 \rightarrow 8\) on \(+\)
\(8 \rightarrow 9\) on \((\)
\(9 \rightarrow 5\) on \(int\)
\(9 \rightarrow 10\) on \(E\)
\(10 \rightarrow 8\) on \(+\)
\(10 \rightarrow 11\) on \()\)
States with actions:
1: \(E \rightarrow int\) on \(\$, +\)
2: accept on \(\$\)
5: \(E \rightarrow int\) on \(),+\)
7: \(E \rightarrow E + (E)\) on \(\$,+\)
11: \(E \rightarrow E + (E)\) on \(),+\)
Parsers represent the DFA as a 2D table similar to table-driven lexical analysis
Rows correspond to DFA states
Columns correspond to terminals and non-terminals
Columns are typically split into:
terminals: action table
non-terminals: goto table
| \(int\) | \(+\) | \((\) | \()\) | \(\$\) | \(E\) | |
|---|---|---|---|---|---|---|
| 0 | s1 | g2 | ||||
| 1 | r(\(E \rightarrow int\)) | r(\(E \rightarrow int\)) | ||||
| 2 | s3 | accept | ||||
| 3 | s4 | |||||
| 4 | s5 | g6 | ||||
| 5 | r(\(E \rightarrow int\)) | r(\(E \rightarrow int\)) | ||||
| 6 | s8 | s7 | ||||
| 7 | r(\(E \rightarrow E + (E)\)) | r(\(E \rightarrow E + (E)\)) | ||||
| 8 | s9 | |||||
| 9 | s5 | g10 | ||||
| 10 | s8 | s11 | ||||
| 11 | r(\(E \rightarrow E + (E)\)) | r(\(E \rightarrow E + (E)\)) |
After a shift or reduce action we rerun the DFA on the entire stack
For each stack element remember which state it transitions to in the DFA
The \(LR\) parser maintains a stack \[\langle sym_1, state_1 \rangle \ldots \langle sym_n, state_n \rangle\] where \(state_k\) is the final state of the DFA on \(sym_1 \ldots sym_k\)
let I = w$ be the initial input
let j = 0
let DFA state 0 be the start state
let stack = <dummy, 0>
repeat
case action[top_state(stack), I[j]] of
shift k: push <I[j++], k>
reduce X -> A:
pop |A| pairs
push <X, goto[top_state(stack), X]>
accept: halt normally
error: halt and report errorCan be used to parse more grammars than \(LL\)
Most programming languages are \(LR\)
\(LR\) parsers can be described as a simple table
There are tools for building the table
Open question: how is the table constructed?
The stack describes the context of the parse
What non-terminal we are looking for
What production right hand side we are looking for
What we have seen so far from the right hand side
Each DFA state describes several such contexts
An \(LR(0)\) item is a production with a “\(\vert\)” somewhere on the right hand side
The items for \(T \rightarrow (E)\) are:
\(T \rightarrow \vert (E)\)
\(T \rightarrow (\vert E)\)
\(T \rightarrow (E \vert )\)
\(T \rightarrow (E)\vert\)
The only item for \(X \rightarrow \epsilon\) is \(X \rightarrow \vert\)
An item \(\langle X \rightarrow \alpha \vert \beta \rangle\) says that
the parser is looking for an \(X\)
it has an \(\alpha\) on top of stack
expects to find a string derived from \(\beta\) next in the input
Notes
\(\langle X \rightarrow \alpha \vert a \beta \rangle\) means that \(a\) should follow – then we can shift it and still have a viable prefix
\(\langle X \rightarrow \alpha \vert \rangle\) means that we could reduce \(X\) – but this is not always a good idea
An \(LR(1)\) item is a pair: \[\langle X \rightarrow \alpha \vert \beta, a \rangle\]
\(X \rightarrow \alpha \beta\) is a production
\(a\) is a terminal (the lookahead terminal)
\(LR(1)\) means one lookahead terminal
\(\langle X \rightarrow \alpha \vert \beta, a \rangle\) describes a context of the parser
We are trying to find an \(X\) followed by an \(a\), and
We have (at least) \(\alpha\) already on top of the stack
Thus, we need to see a prefix derived from \(\beta a\)
The symbol \(\vert\) was used before to separate the stack from the rest of the input.
In items \(\vert\) is used to mark a prefix of a production right hand side: \[\langle X \rightarrow \alpha \vert \beta, a \rangle\]
In both cases, the stack is on the left of \(\vert\)
We add to our grammar a fresh new start symbol \(S\) and a production \(S \rightarrow E\) where \(E\) is the old start symbol
The initial parsing context contains: \[\langle S \rightarrow \vert E, \$ \rangle\]
Trying to find an \(S\) as a string derived from \(E\$\)
The stack is empty
In context containing \[\langle E \rightarrow E + \vert (E), + \rangle\] If “(” follows then we can perform a shift to context containing \[\langle E \rightarrow E + (\vert E) , + \rangle\]
In context containing \[\langle E \rightarrow E + (E) \vert , + \rangle\] We can perform a reduction with \(E \rightarrow E + (E)\), but only if a “\(+\)” follows
Consider the item \[\langle E \rightarrow E + (\vert E), + \rangle\]
We expect a string derived from \(E ) +\)
There are two productions for \(E\)
\(E \rightarrow int\)
\(E \rightarrow E + (E)\)
We describe this by extending the context with two more items:
\(\langle E \rightarrow \vert int, ) \rangle\)
\(\langle E \rightarrow \vert E + (E), ) \rangle\)
The operation of extending the context with items is called the closure operation
Closure(Items) =
repeat
for each [X -> alpha | Y beta, a] in Items
for each production Y -> gamma
for each b in First(beta a)
add [Y -> | gamma, b] to Items
until Items is unchangedConstruct the start context: \(Closure( \{ S \rightarrow E, \$ \})\)
\(\langle S \rightarrow \vert E, \$ \rangle\)
\(\langle E \rightarrow \vert E + (E), \$ \rangle\)
\(\langle E \rightarrow \vert int, \$ \rangle\)
\(\langle E \rightarrow \vert E + (E), + \rangle\)
\(\langle E \rightarrow \vert int, + \rangle\)
We abbreviate as:
\(\langle S \rightarrow \vert E, \$ \rangle\)
\(\langle E \rightarrow \vert E + (E), \$/+ \rangle\)
\(\langle E \rightarrow \vert int, \$/+ \rangle\)
A DFA state is a closed set of \(LR(1)\) items
The start state contains \(\langle S \rightarrow \vert E, \$ \rangle\)
A state that contains \(\langle X \rightarrow \alpha \vert b \rangle\) is labelled with “reduce with \(X \rightarrow \alpha\) on \(b\)”
A state “State” that contains \(\langle X \rightarrow \alpha \vert y \beta, b \rangle\) has a transition labeled \(y\) to a state that contains the items “Transition(State,y)” where \(y\) can be a terminal or non-terminal
Transition(State, y) =
Items = empty set
for each [X -> alpha | y beta, a] in State
add [X -> alpha y | beta, b] to Items
return Closure(Items)Parsing tables (DFA) can be constructed automatically for a CFG
But, we still need to understand the construction to work with parser generators
What kinds of errors can we expect?
If a DFA state contains both \[\langle X \rightarrow \alpha \vert a \beta, b \rangle\] and \[\langle Y \rightarrow \gamma \vert, a \rangle\]
Then on input “\(a\)” we could either
Shift into state \(\langle X \rightarrow \alpha a \vert \beta, b \rangle\)
Reduce with \(Y \rightarrow \gamma\)
This is called a shift-reduce conflict
Typically due to ambiguities in the grammar
Classic example: the dangling else \[S \rightarrow if \; E \; then \; S \; \vert \; if \; E \; then \; S \; else \; S \; \vert \; OTHER\]
Will have a DFA state containing
\(\langle S \rightarrow if \; E \; then \; S \vert, else \rangle\)
\(\langle S \rightarrow if \; E \; then \; S \vert \; else \; S, x \rangle\)
If \(else\) follows then we can shift or reduce
The default behavior of tools is to shift
Consider the ambiguous grammar \[E \rightarrow E + E \; \vert \; E * E \; \vert \; int\]
We will have the states containing
\(\langle E \rightarrow E * \vert E, + \rangle \Rightarrow \langle E \rightarrow E * E \vert, + \rangle\)
\(\langle E \rightarrow \vert E + E, + \rangle \Rightarrow \langle E \rightarrow E \vert + E, + \rangle\)
\(\ldots\)
Again we have a shift/reduce on input \(+\)
We need to reduce (\(*\) binds tighter than \(+\))
Recall solution: declare the precedence of \(*\) and \(+\)
In yacc we can declare precedence and associativity
%left +
%left *Precedence of a rule equals that of its last terminal
Resolve shift/reduce conflict with a shift if:
no precedence declared for either rule or terminal
input terminal has a higher precedence than the rule
the precedences are the same and right associative
Back to the example
\(\langle E \rightarrow E * \vert E, + \rangle \Rightarrow \langle E \rightarrow E * E \vert, + \rangle\)
\(\langle E \rightarrow \vert E + E, + \rangle \Rightarrow \langle E \rightarrow E \vert + E, + \rangle\)
\(\ldots\)
Will choose reduce because precedence of rule \(E \rightarrow E * E\) is higher than of terminal \(+\)
Another example
\(\langle E \rightarrow E + \vert E, + \rangle \Rightarrow \langle E \rightarrow E + E \vert, + \rangle\)
\(\langle E \rightarrow \vert E + E, + \rangle \Rightarrow \langle E \rightarrow E \vert + E, + \rangle\)
\(\ldots\)
Now we have a shift/reduce on input \(+\): we choose redue because \(E \rightarrow E + E\) and \(+\) have the same precedence and \(+\) is left associative
The phrase precedence declaration is misleading
These declarations do not define precedence, they define conflict resolutions
That is, they instruct shift-reduce parsers to resolve conflicts in certain ways – that is not quite the same thing as precedence
If a DFA state contains both \[\langle X \rightarrow \alpha \vert, a \rangle\] and \[\langle Y \rightarrow \beta \vert, a \rangle\] then on “\(a\)” we don not know which production to reduce
This is called a reduce/reduce conflict
Usually due to gross ambiguity in the grammar
Example: \[S \rightarrow \epsilon \; \vert \; id \; \vert \; id \; S\]
There are two parse trees for the string \(id\)
This grammar is better if we rewrite it as \[S \rightarrow \epsilon \; \vert \; id \; S\]
A parser generator automatically contructs the parsing DFA given a context free grammar
Use precedence declarations and default conventions to resolve conflicts
The parser algorithm is the same for all grammars
But, most parser generators do not construct the DFA as described before because the \(LR(1)\) parsing DFA has thousands of states for even simple languages
But, many states are similar: \[\langle E \rightarrow int \vert, \$/+ \rangle \; \text{and} \; \langle E \rightarrow int \vert, )/+ \rangle\]
Idea: merge the DFA states whose items differ only in the lookahead tokens
We say that that such states have the same core
In this example, we obtain \[\langle E \rightarrow int \vert, \$/+/) \rangle\]
Definition: The core of a set of \(LR\) items is the set of first components without the lookahead terminals
Example: the core of \[\left\{ \langle X \rightarrow \alpha \vert \beta, b \rangle, \langle Y \rightarrow \gamma \vert \delta, d \rangle \right\}\] is \[\left\{ X \rightarrow \alpha \vert \beta, Y \rightarrow \gamma \vert \delta \right\}\]
Consider for example the \(LR(1)\) states
Example: the core of \[\begin{aligned} &\left\{ \langle X \rightarrow \alpha \vert, a \rangle, \langle Y \rightarrow \beta \vert, c \rangle \right\}\\ &\left\{ \langle X \rightarrow \alpha \vert, b \rangle, \langle Y \rightarrow \beta \vert, d \rangle \right\} \end{aligned}\]
They have the same core and can be merged
The merged state contains: \[\left\{ \langle X \rightarrow \alpha \vert, a/b \rangle, \langle Y \rightarrow \beta \vert, c/d \rangle \right\}\]
These are called \(LALR(1)\) states
Stands for LookAhead LR
Typically 10 times fewer \(LALR(1)\) states than \(LR(1)\)
Repeat until all states have a distinct core
Choose two distinct states with the same core
Merge the states by creating a new one with the union of all the items
Point edges from the predecessors to the new state
New state points to all previous states
Consider for example the \(LR(1)\) states \[\begin{aligned} &\left\{ \langle X \rightarrow \alpha \vert, a \rangle, \langle Y \rightarrow \beta \vert, b \rangle \right\}\\ &\left\{ \langle X \rightarrow \alpha \vert, b \rangle, \langle Y \rightarrow \beta \vert, a \rangle \right\} \end{aligned}\]
And the merged \(LALR(1)\) state \[\left\{ \langle X \rightarrow \alpha \vert, a/b \rangle, \langle Y \rightarrow \beta \vert, a/b \rangle \right\}\]
Has a new reduce/reduce conflict
In practice such cases are rare
\(LALR\) languages are not natural; they are an efficiency hack on \(LR\) languages
Most reasonable programming languages has an \(LALR(1)\) grammar
\(LALR(1)\) parsing has become a standard for programming languages and for parser generators.
We can now illustrate how semantic actions are implemented for \(LR\) parsing
Keep attributes on the stack:
On shifting \(a\), push the attribute for \(a\) on the stack
On reduce \(X \rightarrow \alpha\)
pop attributes for \(\alpha\)
compute attribute for \(X\)
push it on the stack
Recall the example \[\begin{aligned} E & \rightarrow T + E_1 &\{&E.val = T.val + E_1.val\}\\ & \quad \vert \; T &\{&E.val = T.val\}\\ T & \rightarrow int * T_1 &\{&T.val = int.val + T_1.val\}\\ & \quad \vert \; int &\{&T.val = int.val\}\\ \end{aligned}\]
Consider parsing the string: \(4 * 9 + 6\)
Recall the example \[\begin{aligned} E & \rightarrow T + E_1 &\{&E.val = T.val + E_1.val\}\\ & \quad \vert \; T &\{&E.val = T.val\}\\ T & \rightarrow int * T_1 &\{&T.val = int.val + T_1.val\}\\ & \quad \vert \; int &\{&T.val = int.val\}\\ \end{aligned}\]
Consider parsing the string: \(4 * 9 + 6\)
| String | Action |
| \(\vert int * int + int\) | shift |
| \(int(4) \vert * int + int\$\) | shift |
| \(int(4) * \vert int + int\$\) | shift |
| \(int(4) * int(9) \vert + int\$\) | reduce \(T \rightarrow int\) |
| \(int(4) * T(9) \vert + int\$\) | reduce \(T \rightarrow int * T\) |
| \(T(36) \vert + int\$\) | shift |
| \(T(36) + \vert int\$\) | shift |
| \(T(36) + int(6) \vert\$\) | reduce \(T \rightarrow int\) |
| \(T(36) + T(6) \vert\$\) | reduce \(E \rightarrow T\) |
| \(T(36) + E(6) \vert\$\) | reduce \(E \rightarrow T + E\) |
| \(E(42) \vert\$\) | accept |
Parsing
A solid foundation: context-free grammars
A simple parser: \(LL(1)\)
A more powerful parser: \(LR(1)\)
An efficiency hack: \(LALR(1)\)
\(LALR(1)\) parser generators