Plan for this week
Last week:
- How do we evaluate a program given its AST?
This week:
- How do we convert program text into an AST?
Example: calculator with variables
AST representation:
data Aexpr
= AConst Int
| AVar Id
| APlus Aexpr Aexpr
| AMinus Aexpr Aexpr
| AMul Aexpr Aexpr
| ADiv Aexpr Aexpr
Evaluator:
Using the evaluator:
λ> eval [] (APlus (AConst 2) (AConst 6))
8
λ> eval [("x", 16), ("y", 10)] (AMinus (AVar "x") (AVar "y"))
6
λ> eval [("x", 16), ("y", 10)] (AMinus (AVar "x") (AVar "z"))
*** Exception: Error {errMsg = "Unbound variable z"}
But writing ASTs explicitly is really tedious, we are used to writing programs as text!
We want to write a function that converts strings to ASTs if possible:
For example:
λ> parse "2 + 6"
APlus (AConst 2) (AConst 6)
λ> parse "(x - y) / 2"
ADiv (AMinus (AVar "x") (AVar "y")) (AConst 2)
λ> parse "2 +"
*** Exception: Error {errMsg = "Syntax error"}
Two-step-strategy
How do I read a sentence “He ate a bagel”?
- First split into words:
["He", "ate", "a", "bagel"]
- Then relate words to each other: “He” is the subject, “ate” is the verb, etc
Let’s do the same thing to “read” programs!
Step 1 (Lexing) : From String to Tokens
A string is a list of characters:
First we aggregate characters that “belong together” into tokens (i.e. the “words” of the program):
We distinguish tokens of different kinds based on their format:
- all numbers: integer constant
- alphanumeric, starts with a letter: identifier
+
: plus operator- etc
Step 2 (Parsing) : From Tokens to AST
Next, we convert a sequence of tokens into an AST
- This is hard…
- … but the hard parts do not depend on the language!
Parser generators
- Given the description of the token format generates a lexer
- Given the description of the grammar generates a parser
We will be using parser generators, so we only care about how to describe the token format and the grammar
Lexing
We will use the tool called alex
to generate the lexer
Input to alex
: a .x
file that describes the token format
Tokens
First we list the kinds of tokens we have in the language:
data Token
= NUM AlexPosn Int
| ID AlexPosn String
| PLUS AlexPosn
| MINUS AlexPosn
| MUL AlexPosn
| DIV AlexPosn
| LPAREN AlexPosn
| RPAREN AlexPosn
| EOF AlexPosn
Token rules
Next we describe the format of each kind of token using a rule:
[\+] { \p _ -> PLUS p }
[\-] { \p _ -> MINUS p }
[\*] { \p _ -> MUL p }
[\/] { \p _ -> DIV p }
\( { \p _ -> LPAREN p }
\) { \p _ -> RPAREN p }
$alpha [$alpha $digit \_ \']* { \p s -> ID p s }
$digit+ { \p s -> NUM p (read s) }
Each line consist of:
- a regular expression that describes which strings should be recognized as this token
- a Haskell expression that generates the token
You read it as:
- if at position
p
in the input string - you encounter a substring
s
that matches the regular expression - evaluate the Haskell expression with arguments
p
ands
Regular Expressions
A regular expression has one of the following forms:
[c1 c2 ... cn]
matches any of the charactersc1 .. cn
[0-9]
matches any digit[a-z]
matches any lower-case letter
[A-Z]
matches any upper-case letter[a-z A-Z]
matches any letter
R1 R2
matches a strings1 ++ s2
wheres1
matchesR1
ands2
matchesR2
- e.g.
[0-9] [0-9]
matches any two-digit string
- e.g.
R+
matches one or more repetitions of whatR
matches- e.g.
[0-9]+
matches a natural number
- e.g.
R*
matches zero or more repetitions of whatR
matches
QUIZ
Which of the following strings are matched by [a-z A-Z] [a-z A-Z 0-9]*
?
(A) (empty string)
(B) 5
(C) x5
(D) x
(E) C and D
Back to token rules
We can name some common regexps like:
and write [a-z A-Z] [a-z A-Z 0-9]*
as $alpha [$alpha $digit]*
[\+] { \p _ -> PLUS p }
[\-] { \p _ -> MINUS p }
[\*] { \p _ -> MUL p }
[\/] { \p _ -> DIV p }
\( { \p _ -> LPAREN p }
\) { \p _ -> RPAREN p }
$alpha [$alpha $digit \_ \']* { \p s -> ID p s }
$digit+ { \p s -> NUM p (read s) }
- When you encounter a
+
, generate aPLUS
token - …
- When you encounter a nonempty string of digits, convert it into an integer and generate a
NUM
- When you encounter an alphanumeric string that starts with a letter, save it in an `ID token
Running the Lexer
From the token rules, alex
generates a function alexScan
which
- given an input string, find the longest prefix
p
that matches one of the rules - if
p
is empty, it fails - otherwise, it converts
p
into a token and returns the rest of the string
We wrap this function into a handy function
which repeatedly calls alexScan
until it consumes the whole input string or fails
We can test the function like so:
λ> parseTokens "23 + 4 / off -"
Right [ NUM (AlexPn 0 1 1) 23
, PLUS (AlexPn 3 1 4)
, NUM (AlexPn 5 1 6) 4
, DIV (AlexPn 7 1 8)
, ID (AlexPn 9 1 10) "off"
, MINUS (AlexPn 13 1 14)
]
QUIZ
What is the result of parseTokens "92zoo"
(positions omitted for readability)?
(A) Lexical error
(B) [ID "92zoo"]
(C) [NUM "92"]
(D) [NUM "92", ID "zoo"]
Parsing
We will use the tool called happy
to generate the parser
Input to happy
: a .y
file that describes the grammar
Wait, wasn’t this the grammar?
data Aexpr
= AConst Int
| AVar Id
| APlus Aexpr Aexpr
| AMinus Aexpr Aexpr
| AMul Aexpr Aexpr
| ADiv Aexpr Aexpr
This was abstract syntax
Now we need to describe concrete syntax
- What programs look like when written as text
- and how to map that text into the abstract syntax
Grammars
A grammar is a recursive definition of a set of trees
- each tree is a parse tree for some string
- parse a string
s
= find a parse tree fors
that belongs to the grammar
A grammar is made of:
Terminals: the leaves of the tree (tokens!)
Nonterminals: the internal nodes of the tree
Production Rules that describe how to “produce” a non-terminal from terminals and other non-terminals
- i.e. what children each nonterminal can have:
Aexpr : -- NT Aexpr can have as children:
| Aexpr '+' Aexpr { ... } -- NT Aexpr, T '+', and NT Aexpr, or
| Aexpr '-' AExpr { ... } -- NT Aexpr, T '-', and NT Aexpr, or
| ...
Terminals
Terminals correspond to the tokens returned by the lexer
In the .y
file, we have to declare with terminals in the rules correspond to which tokens from the Token
datatype:
%token
TNUM { NUM _ $$ }
ID { ID _ $$ }
'+' { PLUS _ }
'-' { MINUS _ }
'*' { MUL _ }
'/' { DIV _ }
'(' { LPAREN _ }
')' { RPAREN _ }
Each thing on the left is terminal (as appears in the production rules)
Each thing on the right is a Haskell pattern for datatype
Token
We use
$$
to designate one parameter of a token constructor as the value of that token- we will refer back to it from the production rules
Production rules
Next we define productions for our language:
Aexpr : TNUM { AConst $1 }
| ID { AVar $1 }
| '(' Aexpr ')' { $2 }
| Aexpr '*' Aexpr { AMul $1 $3 }
| Aexpr '+' Aexpr { APlus $1 $3 }
| Aexpr '-' Aexpr { AMinus $1 $3 }
The expression on the right computes the value of this node
$1 $2 $3
refer to the values of the respective child nodes
Example: parsing (2)
as AExpr
:
Lexer returns a sequence of
Token
s:[LPAREN, NUM 2, RPAREN]
LPAREN
is the token for terminal'('
, so let’s pick production'(' Aexpr ')'
Now we have to parse
NUM 2
asAexpr
andRPAREN
as')'
NUM 2
is a token for nonterminalTNUM
, so let’s pick productionTNUM
The value of this
Aexpr
node isAConst 2
, since the value ofTNUM
is2
The value of the top-level
Aexpr
node is alsoAConst 2
(see the'(' Aexpr ')'
production)
QUIZ
What is the value of the root AExpr
node when parsing 1 + 2 + 3
?
Aexpr : TNUM { AConst $1 }
| ID { AVar $1 }
| '(' Aexpr ')' { $2 }
| Aexpr '*' Aexpr { AMul $1 $3 }
| Aexpr '+' Aexpr { APlus $1 $3 }
| Aexpr '-' Aexpr { AMinus $1 $3 }
(A) Cannot be parsed as AExpr
(B) 6
(C) APlus (APlus (AConst 1) (AConst 2)) (AConst 3)
(D) APlus (AConst 1) (APlus (AConst 2) (AConst 3))
Running the Parser
First, we should tell the parser that the top-level non-terminal is AExpr
:
From the production rules and this line, happy
generates a function aexpr
that tries to parse a sequence of tokens as AExpr
We package this function together with the lexer and the evaluator into a handy function
We can test the function like so:
λ> evalString [] "1 + 3 + 6"
10
λ> evalString [("x", 100), ("y", 20)] "x - y"
???
λ> evalString [] "2 * 5 + 5"
???
λ> evalString [] "2 - 1 - 1"
???
Precedence and associativity
The problem is that our grammar is ambiguous!
There are multiple ways of parsing the string 2 * 5 + 5
, namely
APlus (AMul (AConst 2) (AConst 5)) (AConst 5)
(good)AMul (AConst 2) (APlus (AConst 5) (AConst 5))
(bad!)
Wanted: tell happy
that *
has higher precedence than +
!
There are multiple ways of parsing 2 - 1 - 1
, namely
AMinus (AMinus (AConst 2) (AConst 1)) (AConst 1)
(good)AMinus (AConst 2) (AMinus (AConst 1) (AConst 1))
(bad!)
Wanted: tell happy
that -
is left-associative!
How do we communicate precedence and associativity to happy
?
Solution 1: Grammar factoring
We can split the AExpr
non-terminal into multiple “levels”
Aexpr : Aexpr '+' Aexpr2
| Aexpr '-' Aexpr2
| Aexpr2
Aexpr2 : Aexpr2 '*' Aexpr3
| Aexpr2 '/' Aexpr3
| Aexpr3
Aexpr3 : TNUM
| ID
| '(' Aexpr ')'
Intuition: AExpr2
“binds tighter” than AExpr
, and AExpr3
is the tightest
Now I cannot parse the string 2 * 5 + 5
as
AMul (AConst 2) (APlus (AConst 5) (AConst 5))
Why?
Solution 2: Parser directives
This problem is so common that parser generators have a special syntax for it!
What this means:
- All our operators are left-associative
- Operators on the lower line have higher precedence
That’s all folks!