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Java Forum / General / October 2006

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Int->String formatting method

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Deniz Dogan - 26 Sep 2006 08:48 GMT
Hello again! Yes, I have yet another problem for all of you great
programmers in here.

This should actually be a pretty easy task to complete, but somehow the
blood in my veins doesn't seem to flow the right way just now. Here's
the problem:
I have an amount of milliseconds, in the range [0,Integer.MAX_VALUE] and
I want to format it to a String of the format "HH:nn:ss,mmm" where HH
ranges between 00-99, nn between 00-59, ss between 00-59 and mmm between
000-999 (all of the ranges inclusive).
The only thing I can come up with now as a solution is a bunch of
if-then-else statements, but really now, we shouldn't have to go and do
that, do we?

I love stating problems. Looking forward to your help.

/Deniz Dogan
Deniz Dogan - 26 Sep 2006 09:21 GMT
> Hello again! Yes, I have yet another problem for all of you great
> programmers in here.
[quoted text clipped - 13 lines]
>
> /Deniz Dogan

Disregard my post, a friend of mine solved it on four rows of code and
O(1) fashion.
Oliver Wong - 26 Sep 2006 18:46 GMT
>> Hello again! Yes, I have yet another problem for all of you great
>> programmers in here.
[quoted text clipped - 16 lines]
> Disregard my post, a friend of mine solved it on four rows of code and
> O(1) fashion.

   You should have post the solution, in case someone in the future
stumbles upon this post via google and has a similar problem and is looking
for the solution.

   here's my guess at the solution:

public static String toTimeString(long milliseconds) {
 final long MS_PER_SEC = 1000;
 final long MS_PER_MIN = MS_PER_SEC * 60;
 final long MS_PER_HOUR = MS_PER_MIN * 60;
 final long hours = milliseconds / MS_PER_HOUR;
 milliseconds %= MS_PER_HOUR;
 final long minutes = milliseconds / MS_PER_MIN;
 milliseconds %= MS_PER_MIN;
 final long seconds = milliseconds / MS_PER_SEC;
 milliseconds %= MS_PER_SEC;
 StringBuffer sb = new StringBuffer();
 sb.append(hours);
 sb.append(":");
 sb.append(minutes);
 sb.append(":");
 sb.append(seconds);
 sb.append(",");
 sb.append(milliseconds);
 return sb.toString();
}

   - Oliver
Deniz Dogan - 27 Sep 2006 08:57 GMT
>>> Hello again! Yes, I have yet another problem for all of you great
>>> programmers in here.
[quoted text clipped - 45 lines]
>
>    - Oliver

And here's my friend's solution:

public static String formatMillis(int millis) {
    int ihh = millis/3600000; //amount of hours
    int inn = (millis - ihh*3600000) / 60000; //amount of minutes
    int iss = (millis - (ihh*3600000) - (inn * 60000)) / 1000; //amount
of seconds
    int immm = (millis - (ihh*3600000) - (inn * 60000)) - iss*1000;
//amount of milliseconds
    //String representations of the integers:
    String mmm = "" + immm, ss = "" + iss, nn = "" + inn, hh = "" + ihh;
    //Making sure the lengths of the Strings are correct:
    if (mmm.length() == 1) mmm = "00" + mmm;
    else if (mmm.length() == 2) mmm = "0" + mmm;
    if (ss.length() == 1) ss = "0" + ss;
    if (nn.length() == 1) nn = "0" + nn;
    if (hh.length() == 1) hh = "0" + hh;
    return hh + ":" + nn + ":" + ss + "," + mmm;
}

I have a question for you Oliver (or anyone else who'd want to answer
this question), is your solution faster than mine?  I considered using
the modulo operator, but I wasn't sure how the JVM would implement it. I
figured it would just do a lot of division operations until it couldn't
divide it any more, but perhaps it implements it in a different manner?

- Deniz Dogan
Thomas Weidenfeller - 27 Sep 2006 09:41 GMT
> I have a question for you Oliver (or anyone else who'd want to answer
> this question), is your solution faster than mine?

It does not matter. Unless you have evidence that an implementation
really affects an application's performance in an unacceptable way there
are more important criteria to judge an algorithm's implementation. E.g.
maintainability, or robustness regarding change.

> I considered using
> the modulo operator, but I wasn't sure how the JVM would implement it. I
> figured it would just do a lot of division operations until it couldn't
> divide it any more, but perhaps it implements it in a different manner?

That would be an extremely brain-dead thing to do. Sun programmers do a
lot of strange and stupid things in the JDK, but I can't imagine they
would do such a brain-dead modulo operator in the VM. They probably use
the C modulo operator which is probably translated into a modulo
assembler instruction - if the particular CPU provides such an
instruction. Or it is translated into a call into some assembler library
- if an old CPU doesn't provide such an assembler instruction or one
that doesn't fit the JLS modulo definition.

Simplified and broad speaking, the reminder is a by-product of binary
integer division algorithms, and as such usually does take the same or
similar time as a single integer division, whether done in hardware
(CPU' ALU) or software.  Don't they teach binary arithmetic in school
any more?

But again, it doesn't matter. As long as you have no evidence that a
particular operation affects your application in an unacceptable way,
you are wasting your time trying to "optimize" such things.

/Thomas
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Wibble - 27 Sep 2006 12:20 GMT
>> I have a question for you Oliver (or anyone else who'd want to answer
>> this question), is your solution faster than mine?
[quoted text clipped - 29 lines]
>
> /Thomas
That said, Oliver's use of StringBuffer instead of + will
make his code much faster.
Deniz Dogan - 27 Sep 2006 13:49 GMT
[snip]
> That said, Oliver's use of StringBuffer instead of + will
> make his code much faster.

I trust you on that one, but I feel obliged to ask why?

- Deniz Dogan
Oliver Wong - 27 Sep 2006 15:17 GMT
> [snip]
>> That said, Oliver's use of StringBuffer instead of + will
>> make his code much faster.
>
> I trust you on that one, but I feel obliged to ask why?

   First of all, my code is incorrect, because I forgot to check if I need
to prefix '0' before certain of the numbers. For example, I might generate
"1:1:1,1" instead of "01:01:01,0001".

   The exact reason that StringBuffer is faster is that String is difficult
to explain. With a lot of hand waving, the overall reason is String is
immutable, but StringBuilder is not, so String has to do a lot of
contortions to progressively build up the output String, whereas
StringBuilder can perform the String construction in the "obvious" way.

   - Oliver
Deniz Dogan - 27 Sep 2006 15:52 GMT
>> [snip]
>>> That said, Oliver's use of StringBuffer instead of + will
[quoted text clipped - 5 lines]
> need to prefix '0' before certain of the numbers. For example, I might
> generate "1:1:1,1" instead of "01:01:01,0001".

Yes, that's exactly why I didn't quite see how your code was faster.
(And also, the milliseconds should be 001, not 0001 ;-))

>    The exact reason that StringBuffer is faster is that String is
> difficult to explain. With a lot of hand waving, the overall reason is
> String is immutable, but StringBuilder is not, so String has to do a lot
> of contortions to progressively build up the output String, whereas
> StringBuilder can perform the String construction in the "obvious" way.

I think I got it, thanks a lot for your help!
Arne Vajhøj - 30 Sep 2006 21:02 GMT
> That said, Oliver's use of StringBuffer instead of + will
> make his code much faster.

No it does not.

When I test the String code and the StringBuffer code
(fixed to properly add leading zeroes), then the String
code as actually fastest.

That may vary based on platform, Java version etc..

But it is not much faster.

There are simply too few appends on too short
strings to really bring up the StringBuffer advantage.

Arne
Oliver Wong - 27 Sep 2006 15:24 GMT
>> I considered using the modulo operator, but I wasn't sure how the JVM
>> would implement it. I figured it would just do a lot of division
[quoted text clipped - 9 lines]
> provide such an assembler instruction or one that doesn't fit the JLS
> modulo definition.

   I think that modern x86 Intel and AMD chips have a built in modulo
operator. For CPUs which don't have modulo, you could implement it like
this:

C = A % B

temp = A / B
temp = B * A
C = A - temp

   So worst case, an integer modulo would take up 3 "normal" integer
operations worth of time. The usage of "temp" implies an extra register, but
"temp" and "C" could use the same register.

   - Oliver
Arne Vajhøj - 30 Sep 2006 20:58 GMT
>> I have a question for you Oliver (or anyone else who'd want to answer
>> this question), is your solution faster than mine?
[quoted text clipped - 3 lines]
> are more important criteria to judge an algorithm's implementation. E.g.
> maintainability, or robustness regarding change.

Yep.

I would consider the following good enough even though
it is 3 times slower than the proposed code:

    private final static DateFormat df = new
SimpleDateFormat("HH:mm:ss,SSS");
    public static String formatMillis3(int millis) {
        return df.format(new Date(millis -
TimeZone.getDefault().getOffset(millis)));
    }

Arne
Deniz Dogan - 01 Oct 2006 11:49 GMT
>>> I have a question for you Oliver (or anyone else who'd want to answer
>>> this question), is your solution faster than mine?
[quoted text clipped - 17 lines]
>
> Arne

This code was actually what I was looking for in the first place!  I
love pretty Java, no matter how slow it is.


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