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Same code but 10 times the base time [message #134994]

Tue, 26 August 2008 06:17

Kent Tong
Messages: 8
Registered: July 2009

Junior Member

Hi,

I was wondering if it is normal for two identical pieces of code
to have 100 times difference in their base times. For example,
in the code below:

public class Foo {
public static void main(String[] args) {
new Foo().m1();
}

private void m1() {
for (int i = 0; i < 10000; i++) {

}
Bar b = new Bar();
b.m2();
}

}

public class Bar {

public void m2() {
for (int i = 0; i < 10000; i++) {
}
}

}

The m1() has a base time of 0.0299 while m2() has 0.000384. Running it
a few times makes no significant impact.

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Re: Same code but 10 times the base time [message #135007 is a reply to message #134994]

Tue, 26 August 2008 06:23

Kent Tong
Messages: 8
Registered: July 2009

Junior Member

Never mind; found the problem: It's "new Bar()" that is taking up the time.

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Re: Same code but 10 times the base time [message #135032 is a reply to message #134994]

Tue, 26 August 2008 07:42

Vincent MAHE
Messages: 129
Registered: July 2009

Senior Member

Kent Tong a écrit :
> Hi,
>
> I was wondering if it is normal for two identical pieces of code
> to have 100 times difference in their base times. For example,
> in the code below:
>
> public class Foo {
> public static void main(String[] args) {
> new Foo().m1();
> }
>
> private void m1() {
> for (int i = 0; i < 10000; i++) {
>
> }
> Bar b = new Bar();
> b.m2();
> }
>
> }
>
> public class Bar {
>
> public void m2() {
> for (int i = 0; i < 10000; i++) {
> }
> }
>
> }
>
> The m1() has a base time of 0.0299 while m2() has 0.000384. Running it
> a few times makes no significant impact.
>
>
May be the "new Bar()" induces a kind of context change into the JVM?

Vincent

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