On Sunday 26 March 2006 21:10, Geir Magnusson Jr wrote:
> I threw together a toy program when I got back to hotel. I called a
> method that did a minor amount of work (increment the int arg and return
> it). Then I created one that did the same, but then threw an exception.
> Then I did one that threw an exception in which string concat was
> used. Finally, one that thew an exception in which a StringBuffer was
> used.
>
> I found, for 5000000 iterations on a 1.86G Pentium M under WinXP using
> Sun's 1.4.2_10
>
> C:\dev\eclipse\workspace\playspace\java>java -classpath . SpeedTest
> test1 - no exception : 16 millis
> test2 - exception, static string: 11859 millis
> test3 - exception string concat : 13516 millis
> test4 - exception stringbuffert : 14125 millis
This simple example probably exaggerates the effect of the exception - test1
is probably so trivial that the JIT compiler probably optimises it to almost
nothing, while the other tests require e.g. a stack frame in which the
exception can be thrown. It would be interesting to see the results on some
other VMs (including gcj). That being said, these numbers do indicate how
relying on exceptions for normal flow of execution can seriously damage
performance.
> I guess we don't get much benefit of using stringbuffer to replace only
> one concat (or the compiler is smart enough to do that itself anyway...
String concatenations are translated to StringBuffer (or StringBuilder)
operations by the compiler - there is no strcat opcode. If you're building up
a string in stages, e.g.
String s = "foo:";
if (bar) s += "bar";
else s += "baz"
...etc. ...
then it's worth replacing s by a StringBuffer and doing an explicit toString()
at the end:
StringBuffer sb = new StringBuffer("foo");
if (bar) sb.append("bar");
else sb.append("baz");
...etc. ...
String s = sb.toString();
because otherwise the compiler generates code to convert the String to a
StringBuffer and back at every stage. But if you're just writing
"Hello " + name + ", how are you?"
then you save nothing by converting this to
new StringBuffer("Hello ").append(name).append(", how are you?");
because the generated bytecode is the same.
Chris
--
Chris Gray /k/ Embedded Java Solutions BE0503765045
Embedded & Mobile Java, OSGi http://www.k-embedded-java.com/
[EMAIL PROTECTED] +32 3 216 0369
