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Hi,

I haven't programmed in Java for a while and just wrote a program to calc
the odds of winning the British national lottery. BUT it is giving me the
wrong result !!

Here's the relevant bit:

public void DoCalc()
{
float winOdds = 49 * 48 * 47 * 46 * 45 * 44;
System.out.println(winOdds);

float anyOrder = 6 * 5 * 4 * 3 * 2;
System.out.println(anyOrder);

float res = winOdds / anyOrder;

System.out.println(res);
}

It outputs:

1.47841293E9
720.0
2053351.2

But the result should be:

10,068,347,520
720
13,983,816

Don't believe me? Paste the following into your calc program:

(49 * 48 * 47 * 46 * 45 * 44) / (6 * 5 * 4 * 3 * 2) =

Or let google do it:
http://www.google.co.uk/search?hl=en...e+Search&meta=

Am I going crazy? Why doesn't my Java program work?

Please help.

MS

MS wrote:
> I haven't programmed in Java for a while and just wrote a program to
> calc the odds of winning the British national lottery. BUT it is giving
> me the wrong result !!
>
> Here's the relevant bit:
>
> public void DoCalc()
> {
> float winOdds = 49 * 48 * 47 * 46 * 45 * 44;

I am pretty sure you make an integer overflow here.

try:

float winOdds = 49.0 * 48.0 * 47.0 * 46.0 * 45.0 * 44.0;


> System.out.println(winOdds);
>
> float anyOrder = 6 * 5 * 4 * 3 * 2;
> System.out.println(anyOrder);
>
> float res = winOdds / anyOrder;
>
> System.out.println(res);
> }
>
> It outputs:
>
> 1.47841293E9
> 720.0
> 2053351.2
>
> But the result should be:
>
> 10,068,347,520
> 720
> 13,983,816

Arne

MS wrote:
> float winOdds = 49 * 48 * 47 * 46 * 45 * 44;

> 1.47841293E9
> 720.0
> 2053351.2
>
> But the result should be:
>
> 10,068,347,520
> 720
> 13,983,816


What's the value of Integer.MAX_VALUE?

Tom Hawtin

Arne Vajhøj emailed this:
> MS wrote:
>> I haven't programmed in Java for a while and just wrote a program to
>> calc the odds of winning the British national lottery. BUT it is
>> giving me the wrong result !!
>>
>> Here's the relevant bit:
>>
>> public void DoCalc()
>> {
> > float winOdds = 49 * 48 * 47 * 46 * 45 * 44;
>
> I am pretty sure you make an integer overflow here.
>
> try:
>
> float winOdds = 49.0 * 48.0 * 47.0 * 46.0 * 45.0 * 44.0;

Many thanks. What you suggested gave a 'possible loss of precision' error,
when I swapped the floats to doubles I got the correct result.

Strange oversight by the Java designers that Java doesn't assume the .0 at
the end of any real number.

Thanks again.

MS wrote:
> Hi,
>
> I haven't programmed in Java for a while and just wrote a program to
> calc the odds of winning the British national lottery. BUT it is giving
> me the wrong result !!
>
> Here's the relevant bit:
>
> public void DoCalc()
> {
> float winOdds = 49 * 48 * 47 * 46 * 45 * 44;
> // snip
> }
>
> It outputs:
>
> 1.47841293E9
> 720.0
> 2053351.2
>
> But the result should be:
>
> 10,068,347,520
> 720
> 13,983,816
No it shouldn't. float refers to single-precision floating point
scientific values (base two, of course), so ones-digit accuracy may not
possible when working with millions, but the answer would be expressed
(if properly calculated) as something with E7.
>
> Don't believe me? Paste the following into your calc program:
> [snip]
> Am I going crazy? Why doesn't my Java program work?
What Java is doing when it calculates winOdds is it is multiplying the
numbers /as integers/ and then converting to a float. The number
obviously exceeds 2^31-1, so it is rolling over. There are several ways
to avoid this, the easiest being
float winOdds = 1.0f * 49 * 48 * 47 * 46 * 45 * 44;
or
float winOdds = 49.0f * 48 * 47 * 46 * 45 * 44;

The answer is, no, you aren't crazy, but you do need to learn more about
when conversions get applied.

> What's the value of Integer.MAX_VALUE?

Too low for my needs.
Thanks.

Joshua Cranmer emailed this:
> MS wrote:
>> Hi,
>>
>> I haven't programmed in Java for a while and just wrote a program to
>> calc the odds of winning the British national lottery. BUT it is
>> giving me the wrong result !!
>>
>> Here's the relevant bit:
>>
>> public void DoCalc()
>> {
>> float winOdds = 49 * 48 * 47 * 46 * 45 * 44;
>> // snip
>> }
>>
>> It outputs:
>>
>> 1.47841293E9
>> 720.0
>> 2053351.2
>>
>> But the result should be:
>>
>> 10,068,347,520
>> 720
>> 13,983,816
> No it shouldn't. float refers to single-precision floating point
> scientific values (base two, of course), so ones-digit accuracy may not
> possible when working with millions, but the answer would be expressed
> (if properly calculated) as something with E7.
>>
>> Don't believe me? Paste the following into your calc program:
>> [snip]
>> Am I going crazy? Why doesn't my Java program work?
> What Java is doing when it calculates winOdds is it is multiplying the
> numbers /as integers/ and then converting to a float. The number
> obviously exceeds 2^31-1, so it is rolling over. There are several ways
> to avoid this, the easiest being
> float winOdds = 1.0f * 49 * 48 * 47 * 46 * 45 * 44;
> or
> float winOdds = 49.0f * 48 * 47 * 46 * 45 * 44;
>
> The answer is, no, you aren't crazy, but you do need to learn more about
> when conversions get applied.

Many thanks for the explanation.

Cheers.

MS wrote:
> Arne Vajhøj emailed this:
>> MS wrote:
>>> I haven't programmed in Java for a while and just wrote a program to
>>> calc the odds of winning the British national lottery. BUT it is
>>> giving me the wrong result !!
>>>
>>> Here's the relevant bit:
>>>
>>> public void DoCalc()
>>> {
>> > float winOdds = 49 * 48 * 47 * 46 * 45 * 44;
>>
>> I am pretty sure you make an integer overflow here.
>>
>> try:
>>
>> float winOdds = 49.0 * 48.0 * 47.0 * 46.0 * 45.0 * 44.0;
>
> Many thanks. What you suggested gave a 'possible loss of precision'
> error, when I swapped the floats to doubles I got the correct result.
>
> Strange oversight by the Java designers that Java doesn't assume the .0
> at the end of any real number.

Huh? What "real" numbers?

There was absolutely no oversight on the part of the Java designers here. The
compiler worked exactly as documented.

49, 48, 47, et al. are /integers/.
<http://java.sun.com/docs/books/jls/third_edition/html/lexical.html#3.10.1>

not float or double literals
<http://java.sun.com/docs/books/jls/third_edition/html/lexical.html#3.10.2>

You did /integer/ calculation.

Arne pointed out that that cause an /integer/ overflow.

After which you converted the incorrect /integer/ value to float. (Why not
double?)

How is the compiler suppose to "assume the [sic] .0" in an integer?

--
Lew

MS wrote:
> Arne Vajhøj emailed this:
>> MS wrote:
>>> I haven't programmed in Java for a while and just wrote a program to
>>> calc the odds of winning the British national lottery. BUT it is
>>> giving me the wrong result !!
>>>
>>> Here's the relevant bit:
>>>
>>> public void DoCalc()
>>> {
>> > float winOdds = 49 * 48 * 47 * 46 * 45 * 44;
>>
>> I am pretty sure you make an integer overflow here.
>>
>> try:
>>
>> float winOdds = 49.0 * 48.0 * 47.0 * 46.0 * 45.0 * 44.0;
>
> Many thanks. What you suggested gave a 'possible loss of precision'
> error, when I swapped the floats to doubles I got the correct result.
>
> Strange oversight by the Java designers that Java doesn't assume the .0
> at the end of any real number.

Strange oversight by the Java programmers who can't
distinguish between `int' and `double' constants ...

To put it another way: The eventual "target" of an
expression does not influence the way the expression is
calculated. You wrote `float winOdds =' and apparently
expected the right-hand side to somehow magically understand
that it should be evaluated in `float' arithmetic. But the
original is an expression involving `int' values multiplied
together, so it is evaluated according to the rules of `int'.
After the evaluation, the result is converted to `float' for
the assignment -- but by then it's too late, the overflow has
already occurred and the damage has been done.

Perhaps the commonest situation where this misunderstanding
crops up is

double fahrenheit = 212;
double celsius = 5 / 9 * (fahrenheit - 32);

As a test of your newfound understanding of how expressions
are evaluated, can you say what's wrong with this formula?
For extra credit, can you explain why

double celsius = (fahrenheit - 32) * 5 / 9;

works as (presumably) intended?

--
Eric Sosman
lid

Tom Hawtin wrote:
>> What's the value of Integer.MAX_VALUE?

MS wrote:
> Too low for my needs.

Exactly! That was the problem!

--
Lew