Courtesy: my teammate, Ashok...
J S Ashok
04/27/2000 11:11 AM
To: Anurag Panda/India/IBM@IBMIN
cc: Premjith Manapetty/India/IBM@IBMIN, G Rajeev/India/IBM@IBMIN, Rupesh
A Kartha/India/IBM@IBMIN
From: J S Ashok/India/IBM@IBMIN
Subject: Re: puzzling (Document link: Anurag Panda)
Hi Guys,
I think u people must have figured out the reason. Looks very simple.
public static void main(String args[]){
int a = 1;
int b = 1;
a = a++ * ++a;
In this step first a++ so the operators are evaluated. so operand is
taken as 1. Now then a is incremented.
Then for the next operand a is first incremented, so the operands are 1 &
3. So now the operation is carried out.
a = 1 * 3 . Which puts the value of 3 to a.
b = ++b * b++;
In this step first is ++b so the operators are evaluated. so operand
is taken as 2. Now then b is not incremented before
the operation. so ++b and b are 2 operands which makes them 2 and 2. So the
expression becomes b = 2 * 2;
so b is 4.
System.out.println(a);
System.out.println(b);
}
Ashok Jayaraman S
IBM Global Services,
Internet : [EMAIL PROTECTED]
IBMMAIL : J S Ashok/India/IBM
Direct No. : 91-80-526 2355 Extn. : 2501
Fax : 91-80-527 7374
Ramkumar Jeyaraman <[EMAIL PROTECTED]> on 04/27/2000 10:01:43 AM
Please respond to A mailing list about Java Server Pages specification and
reference <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED]
cc: (bcc: Anurag Panda/India/IBM)
Subject: puzzling
Hi,
Sorry to post this silly doubt here, but I need
experts' answers, and I dont find experts anywhere
other than here.
Accoring to Core Java and other books :
1. Unary operators have higher precedence that
multiplication/division operators.
2. Associativity of unary operator is from right to
left.
In light of this, can anybody explain the output of
the following program :
public static void main(String args[]){
int a = 1;
int b = 1;
a = a++ * ++a;
b = ++b * b++;
System.out.println(a);
System.out.println(b);
}
I am unable to figure out the reason for the output.
regards,
ram
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