Your mission, should you choose to accept it..
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I think that since the cube root of Int32.MaxValue is greater than 1024, we can further reduce it using your same technique (10-bit chunks instead of 8-bit chunks):
if (((a | (a << 10) | (a << 20)) & 0xFF7) * ((b | (b << 10) | (b << 20)) & 0xFF7) * ((c | (c << 10) | (c << 20)) & 0xFF7) == 0) { }
That technique is so outdated now. :laugh:
Luc Pattyn [My Articles] Nil Volentibus Arduum
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That technique is so outdated now. :laugh:
Luc Pattyn [My Articles] Nil Volentibus Arduum
Yep. :rolleyes:
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:jig:
Luc Pattyn [My Articles] Nil Volentibus Arduum
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12? that is a lot. Lets use the sign bit now:
if ( (a|-a)&(b|-b)&(c|-c) < 0 ) log("all non-zero");
:-D
Luc Pattyn [My Articles] Nil Volentibus Arduum
Well done. :thumbsup:
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Well done. :thumbsup:
Thanks. The nice thing about this approach is it works for all widths, and for any number of product factors, as long as they are all signed integers. :)
Luc Pattyn [My Articles] Nil Volentibus Arduum
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12? that is a lot. Lets use the sign bit now:
if ( (a|-a)&(b|-b)&(c|-c) < 0 ) log("all non-zero");
:-D
Luc Pattyn [My Articles] Nil Volentibus Arduum
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.. is to write a replacement for
if (a == 0 || b == 0 || c == 0), that - uses at most one comparison. - uses only integer arithmetic. - does not make assumptions about the values of a b and c, except that they are 32-bit 2's complement integers. This entirely useless challenge (is there any other kind?) was inspired by a question on a site that shall not be named, asking for a shorter way to write it. But then people started answering witha * b * c == 0(which is wrong in general, bonus points if you know why) and(a | b | c) == 0which is a nice try but tests whether all of them are zero instead of any of them. That inspired me to search for a solution like that, and I found 2, one of which uses only basic operators. Can you find it?harold aptroot wrote:
entirely useless challenge (is there any other kind?)
Some of these wouldn't be called entirely useless, would they? :) PS: the damn link-paste bug is acting up again.
Luc Pattyn [My Articles] Nil Volentibus Arduum
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harold aptroot wrote:
entirely useless challenge (is there any other kind?)
Some of these wouldn't be called entirely useless, would they? :) PS: the damn link-paste bug is acting up again.
Luc Pattyn [My Articles] Nil Volentibus Arduum
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Aren't they just some other kind of useless? The kind of useless where you won't solve the challenge anyway so why bother.. But you have a point
When everything were useless, then so would be the word itself. :)
Luc Pattyn [My Articles] Nil Volentibus Arduum
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When everything were useless, then so would be the word itself. :)
Luc Pattyn [My Articles] Nil Volentibus Arduum
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12? that is a lot. Lets use the sign bit now:
if ( (a|-a)&(b|-b)&(c|-c) < 0 ) log("all non-zero");
:-D
Luc Pattyn [My Articles] Nil Volentibus Arduum
You're an artist, Luc.
cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP
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Pfft, I don't even need a comparison:
try
{
int x = 1 / a / b / c;
}
catch (DivideByZeroException)
{
// One of them was zero.
}My eyes! It burnssses!
cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP
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You're an artist, Luc.
cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP
I know my bits and bytes, embedded systems do that to a person. :)
Luc Pattyn [My Articles] Nil Volentibus Arduum
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Pfft, I don't even need a comparison:
try
{
int x = 1 / a / b / c;
}
catch (DivideByZeroException)
{
// One of them was zero.
}