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Sun ePractice question 5 loop/post-increment

 
Greenhorn
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https://tmn.sun.com/WLC/GuestLogin?cmd=login1&loginType=1&language=en&id=programmer0000
Question 5
1. public class TeSet {
2. public static void main(String args[]) {
3. int m = 2;
4. int p = 1;
5. int t = 0;
6. for(;p < 5;p++) {<br /> 7. if(t++ > m) {
8. m = p + t;
9. }
10. }
11. System.out.println("t equals " + t);
12. }
13. }
Please tell me where the below logic is wrong. Slashes denote an old value is wiped out and the latest value takes its place:
m=2
p=1/2/3/4
t=0/1/2/3/4
p<5? y<br /> t>m? n. t now 1
p increments to 2
p<5? y<br /> t>m? n. t now 2
p increments to 3
p<5? y<br /> t>m? n. t now 3
p increments to 4
p<5? y<br /> t>m? y.
t increments to 4
m = p+t = 8
The correct answer is 4.
What did I do incorrectly?
 
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Caroline
You have correctly worked it out. The class is outputting t not m.
 
Caroline Bogart
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The devil's in the details... thanks!
 
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U R RIGHT .Dont' Confuse.Try to compile & excecute code when u r confused or have any doubts.
Bye.
Viki.
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Count the flowers of ur garden,NOT the leafs which falls away!
 
Greenhorn
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Ok, so I have a quick, somewhat off the subject question. In the comparison if(t++ > m) is done with m = 0 and t = 0 before the comparison is performed, then it will yield false, but t will be incremented to 1 after the comparison. Is that correct? However, what if the comparison was if(++t > m) with m = 0 and t = 0 before the comparison, then t will be incremented to 1 before they are compared, and the result will be true. Is this correct? Is this a result of operator precedence, or just the way t++ and ++t are defined?
Thanks,
Matt
[This message has been edited by Matt Wheeler (edited November 21, 2001).]
 
Vikrama Sanjeeva
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In this Case:

t will be first check by it's last value.For first time excution it will check by giving value 0.And after comparison has made it will be incremented to 1.

However, In this Case:
t will be first incremented & the it's invremented value will be used in comparison.That is for the first time execution t will be incremented to 1 & then compared with m.
Hope This Answers.
Bye.
Viki.

------------------
Count the flowers of ur garden,NOT the leafs which falls away!
 
Author & Gold Digger
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6
IntelliJ IDE Java
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you're right Matt,
in ++t the comparison yields true.
In fact, unary operator (like ++) have a higher priority than binary ones.
This topic has been discussed many times here, but here it comes:
In t++, the value of t is stored and incremented right away. BUT the value used in the expression is the stored value not the result of the incrementation.
In ++t, the value is first incremented and then used so the result is available right away
HIH
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Valentin Crettaz
Sun Certified Programmer for Java 2 Platform
 
Vikrama Sanjeeva
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Here the ranchers has good deal with increment/decremnet operator.
Try It & Enjoy It. http://www.javaranch.com/ubb/Forum24/HTML/011163.html
Bye.
Viki.
------------------
Count the flowers of ur garden,NOT the leafs which falls away!
 
Don't get me started about those stupid light bulbs.
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