Compilers

Start Lecture #3

Remark: Summary of end of last time.

1. We have a grammar for the simple expressions. It has no ε-productions (good news), but is left recursive (bad news). First eliminate left recursion and then use predictive parsing to write a program (a parser) that constructs a parse tree for any input string (i.e., for any infix expression).
2. But we can do better. We gave an SDT (i.e. gave actions). Again you can eliminate the left recursion (done in the notes). Now, when your parser constructs the parse tree it is more, it has the print statements as additional leaves.
3. Then you just do a lab 1 traversal (post/pre/EulerTour-order) on this enhanced tree with visit(regularNode) is a nop and visit(printNode) just does the print.
4. The result is the postfix for the given infix.
5. Thus, you have constructed a 2 phase compiler (enhanced parser; tree walker) from infix to postfix.
6. In fact for this simple example you can further modify the parser to not actually produce the parse tree (saving memory). However, we won't do this. Compilers for real programming language can't do this spacing saving since the grammars/SDDs/SDTs are not as simple as this example.

2.6: Lexical Analysis

The purpose of lexical analysis is to convert a sequence of characters (the source) into a sequence of tokens. A lexeme is the sequence of characters comprising a single token.

Note that (following the book) we are going out of order. In reality, the lexer operates on the input and the resulting token sequence is the input to the parser. The reason we were able to produce the translator in the previous section without a lexer is that all the tokens were just one character (that is why we had just single digits).

Actually, you never need a lexer. Anything that a lexer can do a parser can do. But lexers are smaller and for software engineering and other reasons are normally used.

2.6.1: Removal of White space and comments

These do not become tokens so that the parser need not worry about them.

2.6.2: Reading ahead

Consider distinguishing x<y from x<=y.

After reading the < we must read another character. If it is y, we have found our token (<). However, we must unread the y so that when asked for the next token, we will start at y. If it is never more than one extra character that must be examined, a single char variable would suffice. A more general solution is discussed in the next chapter (Lexical Analysis).

2.6.3: Constants

This chapter considers only numerical integer constants. They are computed one digit at a time using the formula

    value:=10*value+digit.
  

The value of the constant can be considered the attribute of the token named num. Alternatively, the attribute can be a pointer/index into the symbol table entry for the number (or into a numbers table).

2.6.4: Recognizing identifiers and keywords

The C statement
    sum = sum + x;
contains 6 tokens. The scanner (aka lexer; aka lexical analyzer) will convert the input into
    id = id + id ;
(id standing for identifier).
Although there are three id tokens, the first and second represent the lexeme sum; the third represents x. These two different lexemes must be distinguished.

A related distinction occurs with language keywords, for example then, which are syntactically the same as identifiers. The symbol table is used to accomplishes both distinctions. We assume (as do most modern languages) that the keywords are reserved, i.e., cannot be used as program variables. The we simply initialize the symbol table to contain all these reserved words and mark them as keywords. When the lexer encounters a would-be identifier and searches the symbol table, it finds out that the string is actually a keyword.

As mentioned previously care must be taken when one lexeme is a proper subset of another. Consider
    x<y versus x<=y
When the < is read, the scanner needs to read another character to see if it is an =. But if that second character is y, the current token is < and the y must be pushed back onto the input stream so that the configuration is the same after scanning < as it is after scanning <=.

Also consider then versus thenewvalue, one is a keyword and the other an id.

2.6.5: A lexical analyzer

A Java program is given. The book, but not the course, assumes knowledge of Java.

Since the scanner converts digits into num's we can shorten the grammar above. Here is the shortened version before the elimination of left recursion. Note that the value attribute of a num is its numerical value.

    expr   → expr + term    { print('+') }
    expr   → expr - term    { print('-') }
    expr   → term
    term   → num            { print(num.value) }
  

    expr   → expr + term    { print('+') }
    expr   → expr - term    { print('-') }
    expr   → term
    term   → factor
    factor → ( expr ) | num { print(num,value) }
  

The factor() procedure follows the familiar recursive descent pattern: Find a production with factor as LHS and lookahead in FIRST, then do what the RHS says. That is, call the procedures corresponding to the nonterminals, match the terminals, and execute the semantic actions.

Note that we are now able to consider constants of more than one digit.

2.7: Incorporating a symbol table

The symbol table is an important data structure for the entire compiler. One example of its use is that the semantic actions or rules associated with declarations set the type field of the symbol table entry. Subsequent semantic actions or rules associated with expression evaluation use this type information. For the simple infix to postfix translator (which is typeless), the table is primarily used to store and retrieve <lexeme,token> pairs.

2.7.1: Symbol Table per Scope

There is a serious issue here involving scope. We will learn that lexers are based on regular expressions; whereas parsers are based on the stronger, but more expensive, context-free grammars. Regular expressions are not powerful enough to handle nested scopes. So, if the language you are compiling supports nested scopes, the lexer can only construct the <lexeme,token> pairs. The parser converts these pairs into a true symbol table that reflects the nested scopes. If the language is flat, the scanner can produce the symbol table.

The idea for a language with nested scopes is that, when entering a block, a new symbol table is created. Each such table points to the one immediately outer. This structure supports the most-closely nested rule for symbols: a symbol is in the scope of most-closely nested declaration. This gives rise to a tree of tables.

Interface

• Create table: A new table is created and points to the immediately outer table, which is passed as a argument.
• Insert entry (in the current table).
• Retrieve entry (from the most-closely nested table in which it appears).

Reserved keywords

Simply insert them into the symbol table prior to examining any input. Then they can be found when used correctly and, since their corresponding token will not be id, any use of them where an identifier is required can be flagged. For example a lexer for a C-like language would have insert(int) performed prior to scanning the input.

2.7.2: The Use of Symbol Tables

Below is the grammar for a stripped down example showing nested scopes. The language consists just of nested blocks, a weird mixture of C- and ada-style declarations (specifically, type colon identifier), and trivial statements consisting of just an identifier.

    program → block
    block   → { decls stmts }     -- { } are terminals not actions
    decls   →  decls decl | ε     -- study this one
    decl    → type : id ;
    stmts   → stmts stmt | ε      -- same idea, a list
    stmt    → block | factor ;    -- enables nested blocks
    factor  → id
  

One possible program in this language is

    { int : x ;  float : y ;
      x ; y ;
      { float : x ;
        x ; y ;
      }
      { int : y ;
        x ; y;
      }
      x ; y ;
    }
  

To show that we have correctly parsed the input and obtained its meaning (i.e., performed semantic analysis), we present a translation scheme that digests the declarations and translates the statements so that the above example becomes

{ int; float; { float; float; } { int; int; } }
  

The translation scheme, slightly modified from the book page 90, is shown on the right. First a formatting comment.

This translation scheme looks weird, but is actually a good idea (of the authors): it reconciles the two goals of respecting the ordering and nonetheless having the actions all in one column.

Recall that the placement of the actions within the RHS of the production is significant. The parse tree with its actions is processed in a depth first manner so that the actions are performed in left to right order. Thus an action is executed after all the subtrees rooted by parts of the RHS to the left of the action and is executed before all the subtrees rooted by parts of the RHS to the right of the action.

Consider the first production. We want the action to be executed before processing block. Thus the action must precede block in the RHS. But we want the actions in the right column. So we split the RHS over several lines and place an action in the rightmost column of the line that puts in the right order.

The second production has some semantic actions to be performed at the start of the block, and others to be performed at the bottom.

To fully understand the details, you must read the book; but we can see how it works. A new Env initializes a new symbol table; top.put inserts into the symbol table of the current environment top; top.get retrieves from that symbol table.

In some sense the star of the show is the simple production
    factor → id
together with its semantic actions. These actions look up the identifier in the (correct!) symbol table and print out the type name.

Question: Why do we have the trivial-looking production

    program → block
  

Hard Question: I prefer ada-style declarations, which are of the form
    identifier : type
What problem would have occurred had I done so here and how does one solve that problem?
Answer: Both a stmt and a decl would start with id and thus the FIRST sets would not be disjoint. The result would be that need more than one token lookahead to see when the decls end and the stmts begin. The fix is to left-factor the grammar as we will learn in chapter 4.

2.8: Intermediate Code Generation

2.8.1: Two kinds of Intermediate Representations

There are two important forms of intermediate representations.

• Trees, especially parse trees and syntax trees.
• Linear, especially three-address code

Another (but less common) name for parse tree is concrete syntax tree. Similarly another (also less common) name for syntax tree is abstract syntax tree.

Very roughly speaking, (abstract) syntax trees are parse trees reduced to their essential components, and three address code looks like assembler without the concept of registers.

2.8.2: Construction of (Abstract) Syntax Trees

Remarks:

1. Despite the words below, your future lab assignments will likely not require producing abstract syntax trees. Instead, you will be producing concrete syntax trees (parse trees). I may include an extra-credit part of some labs that will ask for abstract syntax trees.
2. Note however that real compilers do not produce parse trees since such trees are larger and have no extra information that the compiler needs. If the compiler produces a tree (many do), it produces an abstract syntax tree.
3. The reason I will not require your labs to produce the smaller trees is that to do so it is helpful to understand semantic rules and semantic actions, which come later in the course. Of course, authors of real compilers have already completed the course before starting the design so this consideration does not apply to them. :-)

Consider the production

    while-stmt → while ( expr ) stmt ;
  

To generate this while node, we execute

    new While(x,y)
  

Syntax Trees for Statements

The book has an SDD on page 94 for several statements. The part for while reads
  stmt → while ( expr ) stmt₁   { stmt.n = new While(expr.n, stmt₁.n); }
The n attribute gives the syntax tree node.

Representing Blocks in Syntax Trees

Fairly easy

    stmt  → block          { stmt.n = block.n }
    block → { stmts }      { block.n = stmts.n }
  

    while ( x == 5 ) {
       blah
       blah
       more
    }
  

1. The tree for x==5.
2. The tree for blah blah more.

Syntax trees for Expressions

When parsing we need to distinguish between + and * to insure that 3+4*5 is parsed correctly, reflecting the higher precedence of *. However, once parsed, the precedence is reflected in the tree itself (the node for + has the node for * as a child). The rest of the compiler treats + and * largely the same so it is common to use the same node label, say OP, for both of them. So we see
    term → term₁ * factor     { term.n = new Op('*', term₁.n, factor.n); }

Note, however, that the SDD (Figure 2.39) essentially constructs both the parse tree and the syntax tree. That latter is constructed as the attributes in the former.

2.8.3: Static Checking

Static checking refers to checks performed during compilation; whereas, dynamic checking refers to those performed at run time. Examples of static checks include

• Syntactic checks such as avoiding multiple declarations of the same identifier in the same scope.
• Type checks.

      struct symtableType {
        char lexeme[BIGNUMBER];
        int  token;
      } symtable[ANOTHERBIGNUMBER];
    

L-values and R-values

Consider Q := Z; or A[f(x)+B*D] := g(B+C*h(x,y));. I am using [] for array reference and () for function call).

From a macroscopic view, executing either of these assignments has three components.

1. Evaluate the left hand side (LHS) to obtain an l-value.
2. Evaluate the RHS to obtain an r-value.
3. Perform the assignment.

Note the differences between L-values, quantities that can appear on the LHS of an assignment, and and R-values, quantities that can appear only on the RHS.

• An l-value corresponds to an address or a location.
• An r-value corresponds to a value.
• Neither 12 nor s+t can be used as an l-value, but both are legal r-values.

Static checking is used to insure that R-values do not appear on the LHS.

Type Checking

These checks assure that the type of the operands are expected by the operator. In addition to flagging errors, this activity includes

• Coercions. The automatic conversion of one type to another. Later in this course we will employ the function widen(a,t,w) in certain semantic rules. Widen(x,int,double) would for example generate the intermediate code needed to convert x of type int into a quantity of type double.
• Overloading. In Java, Ada, and other languages, the same symbol can have different meanings depending on the types of the operands. Static checks are used to determine the correct operation, or signal an error if none exists

2.8.4: Three-Address Code

These are primitive instructions that have one operator and (up to) three operands, all of which are addresses. One address is the destination, which receives the result of the operation; the other two addresses are the sources of the values to be operated on.

Perhaps the clearest way to illustrate the (up to) three address nature of the instructions is to write them as quadruples or quads.

    ADD        x y z
    MULT       a b c
    ARRAY_L    q r s
    ARRAY_R    e f g
    ifTrueGoto x L
    COPY       r s
  

    x = y + z
    a = b * c
    q[r] = s
    e = f[g]
    ifTrue x goto L
    r = s
  

Translating Statements

We do this and the next section much slower and in much more detail later in the course.

Here is the if example from the book, which is somewhat Java intensive.

    class If extends Stmt {
       Expr E; Stmt S;
       public If(Expr x, Stmt y) { E = x;  S = y;  after = newlabel(); }
       public void gen() {
          Expr n = E.rvalue();
          emit ("ifFalse " + n.toString() + " goto " + after);
          S.gen();
          emit(after + ":");
       }
    }
  

The idea is that we are to translate the statement

    if expr then stmt
  

1. A block of code to compute the expression placing the result into x.
2. The single line   ifFalse x goto after.
3. A block of code for the statement(s) inside the then
4. The label after.

The constructor for an IF node is called with nodes for the expression and statement. These are saved.

When the entire tree is constructed, the main code calls gen() of the root. As we see for IF, gen of a node invokes gen of the children.

Translating Expressions

I am just illustrating the simplest case

    Expr rvalue(x : Expr) {
       if (x is an Id or Constant node) return x;
       else if (x is an Op(op, y, z) node) {
	  t = new temporary;
	  emit string for t = rvalue(y) op rvalue(z);
	  return a new node for t;
       else read book for other cases
  

Better Code for Expressions

So called optimization (the result is far from optimal) is a huge subject that we barely touch. Here are a few very simple examples. We will cover these since they are local optimizations, that is they occur within a single basic block (a sequence of statements that execute without any jumps).

1. For a Java assignment statement x = x + 1; we would generate two three-address instructions

   	temp = x + 1
   	x    = temp
         

   The can be combined into the three-address instruction x = x + 1, providing there are no further uses of the temporary.
2. Common subexpressions occurring in two different expressions, need be computed only once.

      lvalue y
      push 7
      rvalue xx
      *
      push 6
      rvalue z
      rvalue w
      +
      *
      +
      :=
    

      stmt → id := expr
        { stmt.t := 'lvalue' || id.lexime || expr.t || := }
    

      stmt → if expr then stmt1 { out := newlabel();
                               stmt.t := expr.t || 'gofalse' out || stmt1.t || 'label' out
    

      stmt → if
	 expr      { out := newlabel; emit('gofalse', out); }
	 then
         stmt1     { emit('label', out) }
    

      procedure stmt
        integer test, out;
        if lookahead = id then       // first set is {id} for assignment
          emit('lvalue', tokenval);  // pushes lvalue of lhs
          match(id);                 // move past the lhs]
          match(':=');               // move past the :=
          expr;                      // pushes rvalue of rhs on tos
          emit(':=');                // do the assignment (Omitted in book)
        else if lookahead = 'if' then
          match('if');               // move past the if
          expr;                      // pushes boolean on tos
          out := newlabel();
          emit('gofalse', out);      // out is integer, emit makes a legal label
          match('then');             // move past the then
          stmt;                      // recursive call
          emit('label', out)         // emit again makes out legal
        else if ...                  // while, repeat/do, etc
        else error();
      end stmt;
    

       start → list eof
        list → expr ; list
        list →  ε                   // would normally use | as below
        expr → expr + term      { print('+') }
             | expr - term      { print('-'); }
             | term
        term → term * factor    { print('*') }
             | term / factor    { print('/') }
             | term div factor  { print('DIV') }
             | term mod factor  { print('MOD') }
             | factor
      factor → ( expr )
             | id               { print(id.lexeme) }
             | num              { print(num.value) }
    

            start → list eof
    	 list → expr ; list
    	      | ε
    	 expr → term moreterms
        moreterms → + term { print('+') } moreterms
    	      | - term { print('-') } moreterms
    	      | ε
             term | factor morefactors
      morefactors → * factor { print('*') } morefactors
    	      | / factor { print('/') } morefactors
    	      | div factor { print('DIV') } morefactors
    	      | mod factor { print('MOD') } morefactors
    	      | ε
           factor → ( expr )
                  | id               { print(id.lexeme) }
                  | num              { print(num.value) }
    

    term() {
      int t;
      factor();
      // now we should call morefactorsl(), but instead code it inline
      while(true)              // morefactor nonterminal is right recursive
         switch (lookahead) {  // lookahead set by match()
         case '*': case '/': case DIV: case MOD: // all the same
            t = lookahead;     // needed for emit() below
            match(lookahead)   // skip over the operator
            factor();          // see grammar for morefactors
            emit(t,NONE);
            continue;          // C semantics for case
         default:              // the epsilon production
            return;
  

Two Questions

1. How come this compiler was so easy?
2. Why isn't the final exam next week?

One reason is that much was deliberately simplified. Specifically note that

• No real machine code generated (no back end).
• No optimizations (improvement to generated code).
• FIRST sets disjoint.
• No semantic analysis.
• Input language very simple.
• Output language very simple and closely related to input.

Also, I presented the material way too fast to expect full understanding.

Chapter 3: Lexical Analysis

Homework: Read chapter 3.

Two methods to construct a scanner (lexical analyzer).

1. By hand, beginning with a diagram of what lexemes look like. Then write code to follow the diagram and return the corresponding token and possibly other information.
2. Feed the patterns describing the lexemes to a lexer-generator, which then produces the scanner. The historical lexer-generator is Lex; a more modern one is flex.

Note that the speed (of the lexer not of the code generated by the compiler) and error reporting/correction are typically much better for a handwritten lexer. As a result most production-level compiler projects write their own lexers.

3.1: The Role of the Lexical Analyzer

The lexer is called by the parser when the latter is ready to process another token.

The lexer also might do some housekeeping such as eliminating whitespace and comments. Some call these tasks scanning, but others user the term scanner for the entire lexical analyzer.

After the lexer, individual characters are no longer examined by the compiler; instead tokens (the output of the lexer) are used.

3.1.1: Lexical Analysis Versus Parsing

Why separate lexical analysis from parsing? The reasons are basically software engineering concerns.

1. Simplicity of design. When one detects a well defined subtask (produce the next token), it is often good to separate out the task (modularity).
2. Efficiency. With the task separated out, it is easier to apply specialized techniques.
3. Portability. Only the lexer need communicate with the outside.

3.1.2: Tokens, Patterns, and Lexemes

• A token is a <name,attribute> pair. These are what the parser processes. The attribute might actually be a tuple of several attributes or a pointer to a table entry, which itself might contain many components.
• A pattern describes the character strings for the lexemes of the token. For example a letter followed by a (possibly empty) sequence of letters and digits.
• A lexeme for a token is a sequence of characters that matches the pattern for the token.

Note the circularity of the definitions for lexeme and pattern.

Common token classes.

1. One for each keyword. The pattern is trivial.
2. One for each operator or class of operators. A typical class is the comparison operators. Note that these have the same precedence. We might have + and - as the same token, but not + and *.
3. One for all identifiers (e.g. variables, user defined type names, etc).
4. Constants (i.e., manifest constants) such as 6 or hello, but not a constant identifier such as quantum in the Java statement.
   static final int quantum = 3;. There might be one token for integer constants, a different one for real, another for string, etc.
5. One for each punctuation symbol.

3.1.3: Attributes for Tokens

We saw an example of attributes in the last chapter.

For tokens corresponding to keywords, attributes are not needed since the name of the token tells everything. But consider the token corresponding to integer constants. Just knowing that the we have a constant is not enough, subsequent stages of the compiler need to know the value of the constant. Similarly for the token identifier we need to distinguish one identifier from another. The normal method is for the attribute to specify the symbol table entry for this identifier.

We really shouldn't say symbol table. As mentioned above if the language has scoping (nested blocks) the lexer can't construct the symbol table, but just makes a table of <lexeme,token> pairs, which the parser later converts into a proper symbol table (or tree of tables).

Homework: 1.

3.1.4: Lexical Errors

We saw in this movie an example where parsing got stuck because we reduced the wrong part of the input string. We also learned about FIRST sets that enabled us to determine which production to apply when we are operating left to right on the input. For predictive parsers the FIRST sets for a given nonterminal are disjoint and so we know which production to apply. In general the FIRST sets might not be disjoint so we have to try all the productions whose FIRST set contains the lookahead symbol.

All the above assumed that the input was error free, i.e. that the source was a sentence in the language. What should we do when the input is erroneous and we get to a point where no production can be applied?

In many cases this is up to the parser to detect/repair. Sometimes, however, the lexer is stuck because there are no patterns that match the input at this point.

The simplest solution is to abort the compilation stating that the program is wrong, perhaps giving the line number and location where the lexer and/or parser could not proceed.

We would like to do better and at least find other errors. We could perhaps skip input up to a point where we can begin anew (e.g. after a statement ending semicolon), or perhaps make a small change to the input around lookahead so that we can proceed.

3.2: Input Buffering

Determining the next lexeme often requires reading the input beyond the end of that lexeme. For example, to determine the end of an identifier normally requires reading the first whitespace character after it. Also just reading > does not determine the lexeme as it could also be >=. When you determine the current lexeme, the characters you read beyond it may need to be read again to determine the next lexeme.

3.3: Specification of Tokens

The chapter turns formal and, in some sense, the course begins. The book is fairly careful about finite vs infinite sets and also uses (without a definition!) the notion of a countable set. (A countable set is either a finite set or one whose elements can be put into one to one correspondence with the positive integers. That is, it is a set whose elements can be counted. The set of rational numbers, i.e., fractions in lowest terms, is countable; the set of real numbers is uncountable, because it is strictly bigger, i.e., it cannot be counted.) We should be careful to distinguish the empty set φ from the empty string ε. Formal language theory is a beautiful subject, but I shall suppress my urge to do it right and try to go easy on the formalism.

3.3.1: Strings and Languages

We will need a bunch of definitions.

Definition: An alphabet is a finite set of symbols.

Example: {0,1}, presumably φ (uninteresting), {a,b,c} (typical for exam questions), ascii, unicode, latin-1.

Definition: A string over an alphabet is a finite sequence of symbols from that alphabet. Strings are often called words or sentences.

Example: Strings over {0,1}: ε, 0, 1, 111010. Strings over ascii: ε, sysy, the string consisting of 3 blanks.

Definition: The length of a string is the number of symbols (counting duplicates) in the string.

Example: The length of allan, written |allan|, is 5.

Definition: A language over an alphabet is a countable set of strings over the alphabet.

Example: All grammatical English sentences with five, eight, or twelve words is a language over ascii.

Definition: The concatenation of strings s and t is the string formed by appending the string t to s. It is written st.

Example: εs = sε = s for any string s.

We view concatenation as a product (see Monoid in wikipedia http://en.wikipedia.org/wiki/Monoid). It is thus natural to define s⁰=ε and sⁱ⁺¹=sⁱs.

Example: s¹=s, s⁴=ssss.

More string terminology

A prefix of a string is a portion starting from the beginning and a suffix is a portion ending at the end. More formally,

Definitions:

1. A prefix of s is any string obtained from s by removing (possibly zero) characters from the end of s.
2. A suffix is defined analogously.
3. A substring of s is obtained by deleting a prefix and a suffix.

Note: Any prefix or suffix is a substring.

Examples: If s is 123abc, then

1. s itself and ε are each a prefix, suffix, and a substring.
2. 12 are 123a are prefixes and substrings.
3. 3abc is a suffix and a substring.
4. 23a is a substring

Definitions: A proper prefix of s is a prefix of s other than ε and s itself. Similarly, proper suffixes and proper substrings of s do not include ε and s.

Definition: A subsequence of s is formed by deleting (possibly zero) positions from s. We say positions rather than characters since s may for example contain 5 occurrences of the character Q and we only want to delete a certain 3 of them.

Example: issssii is a subsequence of Mississippi.

Note: Any substring is a subsequence.

Homework: 3(a,b,c).

3.3.2: Operations on Languages

Let L and M be two languages over the same alphabet A.

Definition: The union of L and M, written L ∪ M, is the set-theoretic union, i.e., it consists of all words (strings) in either L or M (or both).

Example: Let the alphabet be ascii. The union of {Grammatical English sentences with one, three, or five words} with {Grammatical English sentences with two or four words} is {Grammatical English sentences with five or fewer words}.

Definition: The concatenation of L and M is the set of all strings st, where s is a string of L and t is a string of M.

We again view concatenation as a product and write LM for the concatenation of L and M.

Examples:: Let the alphabet A={a,b,c,1,2,3}. The concatenation of the languages L={a,b,c} and M={1,2} is L∪M={a1,a2,b1,b2,c1,c2}. The concatenation of {aa,b,c} and {1,2,ε} is {aa1,aa2,aa,b1,b2,b,c1,c2,c}.

Definition: As with strings, it is natural to define powers of a language L.
L⁰={ε}, which is not φ.
Lⁱ⁺¹=LⁱL.

Definition: The (Kleene) closure of L, denoted L^* is
L⁰ ∪ L¹ ∪ L² ...

Definition: The positive closure of L, denoted L⁺ is
L¹ ∪ L² ...

Note: Given either closure it is easy to get the other one.

• L^* = {ε} ∪ L⁺
• L⁺ = L L^*

Example: Let both the alphabet and the language L be {0,1,2,3,4,5,6,7,8,9}. More formally, I would say the alphabet A is {0,1,2,3,4,5,6,7,8,9} and the language L is the set of all strings of length one over A. L⁺ gives all unsigned integers, but with some ugly versions. It has 3, 03, 000003.
{0} ∪ ( {1,2,3,4,5,6,7,8,9} ({0,1,2,3,4,5,6,7,8,9}^* ) ) seems better.

In these notes I may write * for ^* and + for ⁺, but that is strictly speaking wrong and I will not do it on the board or on exams or on lab assignments.

Example: {a,b}* is {ε,a,b,aa,ab,ba,bb,aaa,aab,aba,abb,baa,bab,bba,bbb,...}.
{a,b}+ is {a,b,aa,ab,ba,bb,aaa,aab,aba,abb,baa,bab,bba,bbb,...}.
{ε,a,b}* is {ε,a,b,aa,ab,ba,bb,...}.
{ε,a,b}+ is the same as {ε,a,b}*.

The book gives other examples based on L={letters} and D={digits}, which you should read.

3.3.3: Regular Expressions

The idea is that the regular expressions over an alphabet consist of
the alphabet, and expressions using union, concatenation, and *,
but it takes more words to say it right. For example, I didn't include (). Note that (A ∪ B)* is definitely not A* ∪ B* (* does not distribute over ∪) so we need the parentheses.

The book's definition includes many () and is more complicated than I think is necessary. However, it has the crucial advantages of being correct and precise.

The wikipedia entry doesn't seem to be as precise.

I will try a slightly different approach, but note again that there is nothing wrong with the book's approach (which appears in both first and second editions, essentially unchanged).

Definition: The regular expressions and associated languages over an alphabet consist of

1. ε, the empty string; the associated language L(ε) is {ε}, which is not φ.
2. Each symbol x in the alphabet; L(x) is {x}.
3. rs for all regular expressions (REs) r and s; L(rs) is L(r)L(s).
4. r|s for all REs r and s; L(r|s) is L(r) ∪ L(s).
5. r* for all REs r; L(r*) is (L(r))*.
6. (r) for all REs r; L((r)) is L(r).

Parentheses, if present, control the order of operations. Without parentheses the following precedence rules apply.

The postfix unary operator * has the highest precedence. The book mentions that it is left associative. (I don't see how a postfix unary operator can be right associative or how a prefix unary operator such as unary minus could be left associative.)

Concatenation has the second highest precedence and is left associative.

| has the lowest precedence and is left associative.

The book gives various algebraic laws (e.g., associativity) concerning these operators.

As mentioned above, we don't need the positive closure since, for any RE,

    r+ = rr*.
  

Homework: 2(a-d), 4.

Examples
Let the alphabet A={a,b,c}. Write a regular expression representing the language consisting of all words with

1. at least one c.
2. every b followed by a c.
3. the same number of b's as c's

1. (a|b|c)*c(a|b|c)*
2. (a|bc|c)*
3. IMPOSSIBLE, regular expressions can't count (discussed later in the course)!