#1 Members - Reputation: 113
Posted 15 August 2012 - 09:49 PM
[source lang="java"]public class forDemo {// Demonstrates the for looppublic static void main(String[] args) { // TODO Auto-generated method stub int count; for (count = 0; count < 5; count = count + 1); System.out.println( "This is count: " + count); System.out.println( "Done!"); }}[/source]
The author goes on to say that the example is not good programming and that you will not see programmers use that example, you would probably se this instead: count++.
But when I run it with the latter:
[source lang="java"]public class forDemo {// Demonstrates the for looppublic static void main(String[] args) { // TODO Auto-generated method stub int count; for (count = 0; count < 5; count = count++); System.out.println( "This is count: " + count); System.out.println( "Done!"); }}[/source]
it runs "this is count: 0 infinitely
thanks for any help.
#3 Members - Reputation: 1006
Posted 15 August 2012 - 10:26 PM
It makes more sense as below:
[source lang="java"]int count = 0;count++; // count is 1 after this++count; // count is 2 after thisint a = count++; // a is 2, count is 3 after thisint b = ++count; // b is 4, count is 4 after this[/source]
Edited by jefferytitan, 15 August 2012 - 10:42 PM.
#4 Members - Reputation: 329
Posted 15 August 2012 - 10:37 PM
What should happen is count is assigned the value of count, then it is incremented by 1.
count = count++ //1
count = count++ //2
I'm interested as to why this isn't true in java
#5 Members - Reputation: 113
Posted 15 August 2012 - 10:40 PM
[source lang="java"]for (count = 0; count < 5; count++);[/source]
it gives me: This is count: 5
From what the author says it's suppose to display the exact loop as the first example:
[source lang="java"] for (count = 0; count < 5; count = count +1);[/source]
thanks again.
#6 Members - Reputation: 812
Posted 15 August 2012 - 10:46 PM
#7 Members - Reputation: 113
Posted 15 August 2012 - 10:51 PM
for (count = 0; count < 5; count = count + 1); <<<<<semicolon - that's why it didn't work. sorry guys.
It's just count++, not count=count++. It's not better programming per se, just shorter. I don't use it the way you have, but if I recall correctly the return value of count++ is count, whereas the return value of ++count is count+1.
It makes more sense as below:
[source lang="java"]int count = 0;count++; // count is 1 after this++count; // count is 2 after thisint a = count++; // a is 2, count is 3 after thisint b = ++count; // b is 4, count is 4 after this[/source]
So would that be more relevant, er... a better writing of that code? I just don't want to get into the habit of writing code people may frown upon.
#8 Members - Reputation: 113
Posted 15 August 2012 - 10:52 PM
Another thing, if you put a semicolon after a loop statement it turns the loop into a empty loop (since the semicolon by itself is simply the null statement). To fix this problem you need to get rid of the semicolon after the for statement (but only after the for statement).
Yeah I just got that as you posted. Thanks Ruler
#10 Members - Reputation: 1122
Posted 16 August 2012 - 02:50 AM
'count = ++count' would give the expected result, although there is no point in assigning count's value to itself.
Edited by eppo, 16 August 2012 - 02:53 AM.
#11 Members - Reputation: 329
Posted 16 August 2012 - 05:37 AM
The result of the expression 'count++' is the value of count before incrementing it (0). So a repeated 'count = count++' is basically the same as saying 'count = 0'.
'count = ++count' would give the expected result, although there is no point in assigning count's value to itself.
no its not. True, you assigning it 0 but then you increment it by 1.
#12 Members - Reputation: 176
Posted 16 August 2012 - 06:22 AM
For example:
For (i=0;i<5;i++)
{
System.out.print("test: " + i);
}
#13 Members - Reputation: 1301
Posted 16 August 2012 - 06:23 AM
no its not. True, you assigning it 0 but then you increment it by 1.
The typical implementation of a post increment would look like this:
tmp = count;
count = count + 1;
return tmp;
So no, you're not assigning first and then increment, you store the original value, then increment and the actual assignment is a completely different operator that happens last. In fact, one might argue that any compiler optimizing this code in a way that behaves the way you expect could be considered broken. You can't interrupt one operator (++) halfway through to squeeze in a different operator (=) and then go back to finish the first one.
#14 Moderators - Reputation: 6658
Posted 16 August 2012 - 06:25 AM
In Java, both sides of a non-short circuit binary operator are fully evaluated, including side effects, before the operator is applied. In this case it means that count is incremented before assignment occurs, not after.no its not. True, you assigning it 0 but then you increment it by 1.
#15 Members - Reputation: 329
Posted 16 August 2012 - 06:51 AM
In Java, both sides of a non-short circuit binary operator are fully evaluated, including side effects, before the operator is applied. In this case it means that count is incremented before assignment occurs, not after.
no its not. True, you assigning it 0 but then you increment it by 1.
thank you that makes more sense.






