Slideshow:
Grammar
Grammar:
How re-writing rules
defines a grammar -
Example 1
Example of an instantiation of these re-writing rules:
How re-writing rules
defines a grammar -
Example 2
Example of an instantiation of these re-writing rules:
How re-writing rules
defines a grammar -
Example 3
Example of an instantiation of these re-writing rules:
Re-writing instantiation and parse tree
A re-writing instatiation corresponds to a parse tree !
The re-writing instatiation for 1 + 2 + 3:
The corresponding parse tree:
Re-writing instantiation and parse tree
A re-writing instatiation corresponds to a parse tree !
The re-writing instatiation for 1 + 2 + 3:
The corresponding parse tree:
- Grammar:
- A grammar is
defined by a set of
re-writing rules
- A re-writing rule has the following
form:
< A > ::= Re-write_RuleMeaning:
- <A> can be expressed (= replaced by) the right-hand-side (re-write rule)
- A grammar is
defined by a set of
re-writing rules
- Example: re-writing rules
<expr> ::= <term> | <expr> + <term> | <expr> − <term> <term> ::= <factor> | <term> * <factor> | <term> / <factor> <factor> ::= <constant> | ( <expr> )
- Examples: instantiations
of syntax (re-writing) rules
- 3 is an
<expr> because:
<expr> ::= <term> ( <term> ::= <factor> ) ::= <factor> ( <factor> ::= <constant> ) ::= <constant> ::= 3
- 3 + 4 is an
<expr> because:
<expr> ::= <expr> + <term> ( <expr> ::= <term> ) ::= <term> + <term> ( <term> ::= <factor> ) ::= <factor> + <factor> ( <factor> ::= <constant> ) ::= <constant> + <constant> ::= 3 + 4
- 3 + 4 * 5 is an
<expr> because:
<expr> ::= <expr> + <term> ( <expr> ::= <term> ) ::= <term> + <term> ( <term> ::= <factor> ) ::= <factor> + <term> ( <factor> ::= <constant> ) ::= <constant> + <term> ::= 3 + <term> ( <term> ::= <term> * <factor> ) ::= 3 + <term> * <factor> ::= 3 + <factor> * <factor> ::= 3 + <constant> * <constant> ::= 3 + 4 * 5
- 3 is an
<expr> because:
- Fact:
- The order of the instantiation of the syntax rules corresponds uniquely with a parse tree
- Example 1:
1 + 2 + 3
- Instantiations:
<expr> ::= <expr> + <term> ( <expr> ::= <expr> + <term> ) ::= (<expr> + <term>) + <term> ( <expr> ::= <term> ) ::= (<term> + <term>) + <term> ( <term> ::= <factor> ) ::= (<fact> + <fact>) + <fact> ( <factor> ::= <constant> ) ::= (<const> + <const>) + <const> ::= (1 + 2) + 3 The ( ) are not part of the instantiation I put it there to show the associationParse tree:
Notice that:
- The derivation (parse tree)
can enforce
an ordering (grouping/priority) of
syntax elements !!!
- Example:
- According to the
re-write rules, we
must interpret
1 + 2 together
as an <expr>
- So 1 + 2 is evaluated before other associations !!!
- According to the
re-write rules, we
must interpret
1 + 2 together
as an <expr>
- The derivation (parse tree)
can enforce
an ordering (grouping/priority) of
syntax elements !!!
- Instantiations:
- Example 2:
1 + 2 * 3
- Instantiations:
<expr> ::= <expr> + <term> ( <expr> ::= <term> ) ::= <term> + <term> ( <term> ::= <term> * <factor>) ::= <term> + (<term> * <factor>) ( <term> ::= <factor> ::= <fact> + (<fact> * <fact>) ( <fact> ::= <constant> ) ::= <const> + (<const> * <const>) ::= 1 + (2 * 3) The ( ) are not part of the instantiation I put it there to so associationParse tree:
Notice that:
- The derivation (parse tree)
can enforce
an ordering (grouping/priority) of
syntax elements !!!
- Example:
- According to the
re-write rules, we
must interpret
2 * 3 together
as an <term>
- So 2 * 3 is evaluated before other associations !!!
- According to the
re-write rules, we
must interpret
2 * 3 together
as an <term>
- The derivation (parse tree)
can enforce
an ordering (grouping/priority) of
syntax elements !!!
- Instantiations:
- Demo:
- /home/cs554001/demo/Parsing