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Compiler Construction Sunita M. Dol, CSE Dept
Walchand Institute of Technology, Solapur Page 1
Assignment No.9
AIM:
Implement Operator Precedence Parser
THEORY:
The grammars which have the property that no production right side is epsilon or
has two adjacent non-terminals is called an operator grammar.
The following grammar for expressions
E→EAE | (E) |-E | id
A → + | - | * | / | ↑
is not an operator grammar, because the right side EAE has two (in fact three)
consecutive nonterminal, However, if we substitute for A each of its alternatives,
we obtain the following operator grammar;
E→E+E | E-E | E*E | E/E | E↑E | (E) |-E | id
This parser relies on the following three precedence relations: ⋖, ≐, ⋗
a ⋖ b This means a “yields precedence to” b.
a ⋗ b This means a “takes precedence over” b.
a ≐ b This means a “has precedence as” b.
Figure 1.Operator precedence relation table for grammar E->E+E/E*E/id
Compiler Construction Sunita M. Dol, CSE Dept
Walchand Institute of Technology, Solapur Page 2
There is not given any relation between id and id as id will not be compared and
two variables cannot come side by side. There is also a disadvantage of this table
as if we have n operators than size of table will be n*n and complexity will be
0(n2
). In order to increase the size of table, use operator function table.
The operator precedence parsers usually do not store the precedence table with the
relations; rather they are implemented in a special way. Operator precedence
parsers use precedence functions that map terminal symbols to integers, and so
the precedence relations between the symbols are implemented by numerical
comparison. The parsing table can be encoded by two precedence
functions f and g that map terminal symbols to integers. We select f and g such
that:
1. f(a) < g(b) whenever a is precedence to b
2. f(a) = g(b) whenever a and b having precedence
3. f(a) > g(b) whenever a takes precedence over b
Example – Consider the following grammar:
E -> E + E/E * E/( E )/id
The directed graph representing the precedence function:
Compiler Construction Sunita M. Dol, CSE Dept
Walchand Institute of Technology, Solapur Page 3
Since there is not any cycle in the graph so we can make function table:
PROGRAM:
INPUT AND OUTPUT:
Operator precedence parser
==================
Enter the string(Use i for an identifier)i*(i+i)-i/(i^i)
$ i*(i+i)-i/(i^i)$
$i *(i+i)-i/(i^i)$
$E *(i+i)-i/(i^i)$
Compiler Construction Sunita M. Dol, CSE Dept
Walchand Institute of Technology, Solapur Page 4
$E* (i+i)-i/(i^i)$
$E*( i+i)-i/(i^i)$
$E*(i +i)-i/(i^i)$
$E*(E +i)-i/(i^i)$
$E*(E+ i)-i/(i^i)$
$E*(E+i )-i/(i^i)$
$E*(E+E )-i/(i^i)$
$E*(E )-i/(i^i)$
$E*E -i/(i^i)$
$E -i/(i^i)$
$E- i/(i^i)$
$E-i /(i^i)$
$E-E /(i^i)$
$E-E/ (i^i)$
$E-E/( i^i)$
$E-E/(i ^i)$
$E-E/(E ^i)$
$E-E/(E^ i)$
$E-E/(E^i )$
$E-E/(E^E )$
$E-E/(E )$
$E-E/E $
$E-E $
$E $
Accepted
Compiler Construction Sunita M. Dol, CSE Dept
Walchand Institute of Technology, Solapur Page 5
CONCLUSION:
Operator Precedence Parser which have the property that no production right side
is epsilon or has two adjacent non-terminals is implemented.
REFERENCES:
 Compilers - Principles, Techniques and Tools - A.V. Aho, R. Shethi and J. D.
Ullman (Pearson Education)
 https://www.geeksforgeeks.org/theory-computation-operator-grammar-
precedence-parser/

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Assignment9

  • 1. Compiler Construction Sunita M. Dol, CSE Dept Walchand Institute of Technology, Solapur Page 1 Assignment No.9 AIM: Implement Operator Precedence Parser THEORY: The grammars which have the property that no production right side is epsilon or has two adjacent non-terminals is called an operator grammar. The following grammar for expressions E→EAE | (E) |-E | id A → + | - | * | / | ↑ is not an operator grammar, because the right side EAE has two (in fact three) consecutive nonterminal, However, if we substitute for A each of its alternatives, we obtain the following operator grammar; E→E+E | E-E | E*E | E/E | E↑E | (E) |-E | id This parser relies on the following three precedence relations: ⋖, ≐, ⋗ a ⋖ b This means a “yields precedence to” b. a ⋗ b This means a “takes precedence over” b. a ≐ b This means a “has precedence as” b. Figure 1.Operator precedence relation table for grammar E->E+E/E*E/id
  • 2. Compiler Construction Sunita M. Dol, CSE Dept Walchand Institute of Technology, Solapur Page 2 There is not given any relation between id and id as id will not be compared and two variables cannot come side by side. There is also a disadvantage of this table as if we have n operators than size of table will be n*n and complexity will be 0(n2 ). In order to increase the size of table, use operator function table. The operator precedence parsers usually do not store the precedence table with the relations; rather they are implemented in a special way. Operator precedence parsers use precedence functions that map terminal symbols to integers, and so the precedence relations between the symbols are implemented by numerical comparison. The parsing table can be encoded by two precedence functions f and g that map terminal symbols to integers. We select f and g such that: 1. f(a) < g(b) whenever a is precedence to b 2. f(a) = g(b) whenever a and b having precedence 3. f(a) > g(b) whenever a takes precedence over b Example – Consider the following grammar: E -> E + E/E * E/( E )/id The directed graph representing the precedence function:
  • 3. Compiler Construction Sunita M. Dol, CSE Dept Walchand Institute of Technology, Solapur Page 3 Since there is not any cycle in the graph so we can make function table: PROGRAM: INPUT AND OUTPUT: Operator precedence parser ================== Enter the string(Use i for an identifier)i*(i+i)-i/(i^i) $ i*(i+i)-i/(i^i)$ $i *(i+i)-i/(i^i)$ $E *(i+i)-i/(i^i)$
  • 4. Compiler Construction Sunita M. Dol, CSE Dept Walchand Institute of Technology, Solapur Page 4 $E* (i+i)-i/(i^i)$ $E*( i+i)-i/(i^i)$ $E*(i +i)-i/(i^i)$ $E*(E +i)-i/(i^i)$ $E*(E+ i)-i/(i^i)$ $E*(E+i )-i/(i^i)$ $E*(E+E )-i/(i^i)$ $E*(E )-i/(i^i)$ $E*E -i/(i^i)$ $E -i/(i^i)$ $E- i/(i^i)$ $E-i /(i^i)$ $E-E /(i^i)$ $E-E/ (i^i)$ $E-E/( i^i)$ $E-E/(i ^i)$ $E-E/(E ^i)$ $E-E/(E^ i)$ $E-E/(E^i )$ $E-E/(E^E )$ $E-E/(E )$ $E-E/E $ $E-E $ $E $ Accepted
  • 5. Compiler Construction Sunita M. Dol, CSE Dept Walchand Institute of Technology, Solapur Page 5 CONCLUSION: Operator Precedence Parser which have the property that no production right side is epsilon or has two adjacent non-terminals is implemented. REFERENCES:  Compilers - Principles, Techniques and Tools - A.V. Aho, R. Shethi and J. D. Ullman (Pearson Education)  https://www.geeksforgeeks.org/theory-computation-operator-grammar- precedence-parser/