zizi21
asked on
Java : Urgent - Unary
Hi,
Sorry for the english. Sitting for exam so a bit panic. pls help...
I have a program like below:
public static void main(String [] args){
int x=3;
int y;
y=x++ % ++x;
System.out.println(y);
System.out.println(x);
}
}
When i ran y is 3 and x is 5. I read from book, it says that unary operators are evaluated from right to left..
So, if this is the case, ++x is evaluated first ( x=4) , then , x++, so, x=4, y=0 and not 3...pls clarify
Sorry for the english. Sitting for exam so a bit panic. pls help...
I have a program like below:
public static void main(String [] args){
int x=3;
int y;
y=x++ % ++x;
System.out.println(y);
System.out.println(x);
}
}
When i ran y is 3 and x is 5. I read from book, it says that unary operators are evaluated from right to left..
So, if this is the case, ++x is evaluated first ( x=4) , then , x++, so, x=4, y=0 and not 3...pls clarify
Last Comment
Mick Barry
% is not a unary operator
y = 3 % 5; // = 3
ie. they are evaluated left to right
ASKER
but x++ is unary
yes x++ is evaluated first -> 3
then ++x -> 5
then ++x -> 5
ASKER
or..do you mean in precedence, postfix is done before prefix?thanks
no, when evaluating "a % b", a is evaluated before b
ASKER
i was reading the text book, it says for precedence...
postfix right to left
prefix,unary, right to left
binary operators left to right..
this is so confusning :-(
postfix right to left
prefix,unary, right to left
binary operators left to right..
this is so confusning :-(
ASKER
what do you do they mean right to left then ?
yes that is correct, but the binary operator gets evaluated before the unary operators do
so yes the unary operators are evaluated right to left,
and the % is evaluated left to right, thus evaluating x++ first
and the % is evaluated left to right, thus evaluating x++ first
ASKER
if its like this..
y=++a % b++;
can i say that b gets evaluated, then a gets evaluated...and lastly for % , we do a%b ??
sorry, i am still confused...pls help
y=++a % b++;
can i say that b gets evaluated, then a gets evaluated...and lastly for % , we do a%b ??
sorry, i am still confused...pls help
> y=++a % b++;
1. ++a is evaluated
2. b++ is evaluated
3. (result of 1.) % (result of 2.) is evaluated
1. ++a is evaluated
2. b++ is evaluated
3. (result of 1.) % (result of 2.) is evaluated
ASKER
got it , but why do they say , that for postfix has higher precedence, does it mean, that if i have postfix, then i evaluate right to left...
for example
++a ++b ++c
1) c is evaluated first
2) b is evaluated second
3) a is evaluated third
for example
++a ++b ++c
1) c is evaluated first
2) b is evaluated second
3) a is evaluated third
right to left relates to how the unary operator is evaluated, it doesn't affect how the rest of the expression is evaluated.
ASKER
sorry,,..how does unary operator gets evaluated then ?
http://java.sun.com/docs/books/tutorial/java/nutsandbolts/operators.html
so in your case the unary ops get evaluated first because they have higer precedence.
then the % gets evaluated
so in your case the unary ops get evaluated first because they have higer precedence.
then the % gets evaluated
ASKER
hi, i understand : y=x++ % ++x;
but it says that postfix has higher precedence...
but it says that postfix has higher precedence...
ASKER
Operators Precedence
postfix expr++ expr--
unary ++expr --expr +expr -expr ~ !
multiplicative * / %
additive + -
shift << >> >>>
postfix expr++ expr--
unary ++expr --expr +expr -expr ~ !
multiplicative * / %
additive + -
shift << >> >>>
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ASKER
got it, thank you very much..
Java
Java is a platform-independent, object-oriented programming language and run-time environment, designed to have as few implementation dependencies as possible such that developers can write one set of code across all platforms using libraries. Most devices will not run Java natively, and require a run-time component to be installed in order to execute a Java program.
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