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  A tricky question  (Read 2722 times)
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Offline zingbat

Senior Devvie


Medals: 1


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« Posted 2005-12-21 18:06:48 »

After executing the following instruction:

long time = 60 * 1000000000; // 60 seconds in nano-second units

What will be the value of the variable time ?
Offline Riven
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« Reply #1 - Posted 2005-12-21 18:12:30 »

Your doing:

long x = (long)(int*int);



Do:

long x = long * long;



To prevent overflow of your ints

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Offline zingbat

Senior Devvie


Medals: 1


Java games rock!


« Reply #2 - Posted 2005-12-21 21:30:54 »

Your doing:
long x = (long)(int*int);
Do:
long x = long * long;
To prevent overflow of your ints

I was expecting a newbie to fall for it.  Grin

Anyway you are absolutly right. This is what i used:

long time = 60L * 1000000000L;

But just as a curiosity C++ also messes up with our intuition on this one.
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Offline Kommi

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« Reply #3 - Posted 2005-12-21 21:33:55 »

Oh crap, I better go change my code then Smiley

Kommi
Offline Mr_Light

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Medals: 1


shiny.


« Reply #4 - Posted 2005-12-22 09:29:41 »

well the 2nd number defaults to Long doesn't it?  Kiss

It's harder to read code than to write it. - it's even harder to write readable code.

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Offline Riven
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Hand over your head.


« Reply #5 - Posted 2005-12-22 09:52:13 »

1000000000 = fits perfectly in 32 bits
4294967296 = max for unsigned 32 bits
2147483647 = max for signed 32 bits

And long values *never* default to long.

6000000000 = invalid, compile-time error
6000000000L = valid

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Offline Mr_Light

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Medals: 1


shiny.


« Reply #6 - Posted 2005-12-23 08:10:32 »

6000000000 = invalid, compile-time error
6000000000L = valid
ah yes, horray for compile time checking.

It's harder to read code than to write it. - it's even harder to write readable code.

The gospel of brother Riven: "The guarantee that all bugs are in *your* code is worth gold." Amen brother a-m-e-n.
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