Arithmetic Expressions

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4.3 Arithmetic Expressions
In this section, we shall introduce context-free grammars to
generate a class of arithmetic expressions with operators , ,
,  and ^. Especially, we have to handle the associativities
and precedences of the operators.
For the associativities,
the operators  and  are associative,
the operators  and  are left associative, and
The exponential operator ^ is right associative.
Example 12 :
12 + 4 + 3 = ( ( 12 + 4 ) + 3 ) = ( 12 + ( 4 + 3 ) )
12  4  3 = ( ( 12  4 )  3 ) = ( 12  ( 4  3 ) )
12  4  3 = ( ( 12  4 )  3 )  ( 12  ( 4  3 ) )
12  4  3 = ( ( 12  4 )  3 )  ( 12  ( 4  3 ) )
12 ^ 4 ^ 3 = ( 12 ^ ( 4 ^ 3 ) ) = ( 12 ^ 64 )
 (12 ^ 4) ^ 3 = 12 ^ 12
The following operators takes precedence in decreasing order.
 ( unary minus )
^ ( exponent )
, 
, .
Example 13 :
12  2 + 15  3  4 ^ 3 = ( 12  2 ) + ( (15  3)  (4 ^ 3) )
Left associativity
Example 14: The following grammar as in the example 10 is
ambiguous.
E  E + E | E  E | (E) | <integer>
In order to eliminate ambiguity, we impose left associativities
for the operators + and . After introduce a new variable, say
<term>, we have a new grammar as follows.
E  E + <term> | E  <term> | <term>
<term>  (E) | <integer>
Note : Sometimes we need to apply the theorem 4 to revise the
above left recursive productions to right recursive productions.
The unique parse tree for the sentence 7 + 4  6 + 5 is as follows.
E
E
E
+
<term>
E
+
<term>

<term>
<integer>
*
5
<integer>
*
<term>
<integer>
*
<integer>
4
6
*
7
The parse tree interprets the expression 7+46+5 as (((7+4) 6)+5).
Right associativity
Example 15: The following grammar generating expressions
with operator ^ is ambiguous.
E  E ^ E | (E) | <integer>
The leftmost derivation for 2^3^2 is as follows.
E  E ^ E * 2 ^ E  2 ^ E ^ E * 2 ^ 3 ^ E * 2 ^ 3 ^ 2
The rightmost derivation for 2^3^2 is as follows.
E  E ^ E * E ^ 2  E ^ E ^ 2 * E ^ 3 ^ 2 * 2 ^ 3 ^ 2
Hence, the sentence 2^3^2 has 2 different parse trees as follows.
E
E
E
^
E
E
^
E
<integer>
* <integer> <integer>
2
*
*
3
2
E
<integer>
*
E
^
^
E
E
<integer>
<integer> *
2
*
2
3
The left tree interprets 2^3^2 as 2^(3^2)=2^9 = 512.
The right tree interprets 2^3^2 as (2^3)^2 = 8^2 = 64, a wrong result.
In order to eliminate ambiguity, we impose right associativity for
the operator ^. After introduce a new variable, say <expon>, we
have a new grammar as follows.
Example 16: The following grammar generates expressions for
operator ^ to possess right associativity.
E  <expon> ^ E | <expon>
<expon>  (E) | <integer>
Consider the leftmost derivation and the rightmost derivation for the
sentence 2^3^2.
E  <expon> ^ E * 2 ^ E  2 ^ <expon> ^ E * 2 ^ 3 ^ E
* 2 ^ 3 ^ <expon> * 2^3^2
E  <expon> ^ E  <expon> ^ <expon>^E  <expon> ^
<expon> ^ <expon> * <expon> ^<expon> ^ 2 * 2 ^ 3 ^ 2
Hence, the sentence 2^3^2 has a unique parse tree as follows.
E
<expon> ^
E
<expon> ^
E
<integer>
* <integer> <expon>
2
*
<integer>
3
*
2
Precedence
Both the operators + and  are (left) associative. The operator 
has higher precedence than the operator +. Consider the
grammar in the following example 17. Although it generates
expressions for operators to posses left associativities, it is not
correct because of ignoring the precedence rule.
Example 17: Consider the grammar
E  E + <term> | E  <term> | <term>
<term>  (E) | <integer>
Consider the leftmost derivation and the rightmost derivation for the
sentence 2 + 3  2.
The leftmost derivation for 2 + 3  2 is as follows.
E  E  <term>  E + <term>  <term>  <term> +
<term>  <term> * 2 + 3  2
The rightmost derivation for 2 + 3  2 is as follows.
E  E  <term> * E  2  E + <term>  2 <term> +
<term>  2 * 2 + 3  2
There is a unique parse tree for the derivation of 2 + 3  2 as
follows.
But, the sentence 2 + 3  2 is interpreted as (2+3)  2 = 10, which
is not correct.
E
E
E
<term>
<integer>
*
2

+ <term>
<integer>
*
3
<term>
<integer>
*
2
In order to impose a precedence rule for the operators, we need
to introduce a variable for each precedence level of operators.
the parentheses and a single number have the highest
precedence, let it be the variable <basic>. We have productions:
<basic>  (E) | <integer>
Next, should be the operator , let it be the variable <prod>. We
have productions:
<prod>  <prod>  <basic> | <basic>
Then consider the operator +, let it be the variable E. We have
productions:
E  E + <prod> | <prod>
Example 18: The following grammar generates expressions for
operators + and  to obey the precedence rule.
E  E + <prod> | <prod>
<prod>  <prod>  <basic> | <basic>
<basic>  (E) | <integer>
The rightmost derivation for the sentence 7 + 4  6 + 5 is as follows.
E  E + <prod>  E + <prod> + <basic> * E + <prod> + 5
 E + <prod>  <basic> + 5 * E + <prod>  6 + 5  E +
<basic>  6 + 5 * E + 4  6 + 5  <prod> + 4  6 + 5 
<basic> + 4  6 + 5 * 7 + 4  6 + 5
The unique parse tree for the sentence 7 + 4  6 + 5 is as follows.
E
E
E
+
+
<prod>
<prod>
<basic>
<prod>  <basic> <integer>
*
*
<basic> <basic>
<integer>
5
*
<integer> <integer>
6
<prod>
*
7
*
4
The parse tree interprets 7 + 4  6 + 5 as ((7 + (4  6)) + 5).
Example 19: The following grammar generates expressions for
operators +, , ,  and ^ to obey the precedence rules and
associativities.
E  E + <prod> | E  <prod> | <prod>
<prod>  <prod>  <expon> | <prod>  <expon> | <expon>
<expon>  <item> ^ <expon> | <item>
<item>  <item> | <basic>
<basic>  (E) | <integer>
[[ right recursive productions : <expon>  <item> ^ B | B
B  B ^ <item> | <item> ]]
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