1. Evaluation Expression
Steve Paks

2. Curriculum Model
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CS
Here I am
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CS
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Java Language Base
CS : Computer
Science
A : Application

3. EVALUATION OF
EXPRESSIONS
• Evaluation expression
– We will pick the first one among the following answers.
x = a/b-c+d*e-a*c
a=4, b=c=2, d=e=3
((4/2)-2)+(3*3)-(4*2) = 0+9-8 = 1
or
(4/2-2+3))*(3-4)*2 = (4/3)*(-1)*2 = -2.6666…
– The same precedence have left-to-right associativity.
a*b/c%d/e is equivalent to ((((a*b)/c)%d)/e)

4. EVALUATION OF
EXPRESSIONS(Cont’d)
• Evaluation Postfix Expression
– The standard way of expression is infix notation.
– But, compiler does not use it.
Infix
Postfix
2+3*4
a*b+5
(1+2)*7
a*b/c
((a/(b-c+d))*(e-a)*c
a/b-c+d*e-a*c
234*+
ab*5+
12+7*
ab*c/
abc–d+/ea-*c*
ab/c–de*+ac*-

5. EVALUATION OF
EXPRESSIONS(Cont’d)
• Evaluation Postfix Expression(Cont’d)
– To evaluate an expression
Postfix Expression 62/3-42*+
Token
6
2
/
3
4
2
*
+
Stack
[0]
6
6
6/2
6/2
6/2-3
6/2-3
6/2-3
6/3-3
6/2-3+4*2
Top
[1]
[2]
2
3
4
4
4*2
2
0
1
0
1
0
1
2
1
0

6. EVALUATION OF
EXPRESSIONS(Cont’d)
• Infix to Postfix
– Algorithm
1. Fully parenthesize the expression.
2. Move all binary operators
so that they replace their corresponding right parentheses.
3. Delete all parentheses.
((((a/b)-c)+(d*e))-a*c)) → ((((ab/-c)+(d*e))-a*c))
((((ab/-c)+(d*e))-a*c)) → ((((ab/c-+(d*e))-a*c))
((((ab/c-+(d*e))-a*c)) → ((((ab/c-+(de*)-a*c))
((((ab/c-+(de*)-a*c)) → ((((ab/c-(de*+-a*c))
((((ab/c-(de*+-a*c)) ((((ab/c-(de*+-ac*)
((((ab/c-(de*+-ac*) ((((ab/c-(de*+ac*ab/c-de*+ac*-

7. EVALUATION OF
EXPRESSIONS(Cont’d)
• Infix to Postfix(Cont’d)
– Example 1
Token
a
+
b
*
c
eos
Stack
[0]
+
+
+
+
Top
[1]
*
*
Output
-1
0
0
1
1
-1
a
a
ab
ab
abc
abc*-
[2]

8. EVALUATION OF
EXPRESSIONS(Cont’d)
• Infix to Postfix(Cont’d)
– Example 2
Token
a
*
(
b
+
c
)
*
d
eos
Stack
[0]
*
*
*
*
*
*
*
*
Top
[1]
(
(
(
(
Output
-1
0
1
1
2
2
0
0
0
0
a
a
a
ab
ab
abc
abc+
abc+*
abc+*d
abc+*d*
[2]
+
+