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you should include iostream.h likewise
<font color=blue>#include</font> <iostream>
and not
<font color=blue>#include</font> <iostream.h>
Nibu thomas
Software Developer
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As someone else said, iostream.h is not standard C++. You also need to put using namespace std; at the top of the code, as standard C++ libraries use the std namespace.
Christian Graus - Microsoft MVP - C++
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Hi All,
Is there a fast way of checking to see if an int is a power of 2 (i.e. 2, 4, 8, 16, 1024)? I know the long way using a loop, but I was hoping for a simple way of doing this.
regards,
Rich
"Programming today is a race between software engineers striving to build bigger and
better idiot-proof programs, and the Universe trying to produce bigger and better idiots.
So far the Universe is winning." -- Rich Cook
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Thanks 
"Programming today is a race between software engineers striving to build bigger and
better idiot-proof programs, and the Universe trying to produce bigger and better idiots.
So far the Universe is winning." -- Rich Cook
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I think you will have problems when comparinf the floats, due to the precision.
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true, but that was a starting point.
it would be better to use double at least, but still remains the precision problem.
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2^3 = 8
but
floor(sqrt(8)) != sqrt(8)?
Steve
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Modified version is still wrong - It will treat any integer as a power of 2.
Steve
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i did not pretend to modified my post for a much working stuff...
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if a number is a power of 2 than looking at the bits that composed that value must be all '0' but only one '1'.
So, if int is a 32 bit value, you can look for 31 '0' and 1 '1'.
But you need olso a loop! But it is quite fast.
Let me know if here is a faster bit function to count the ones/zeros
Bye
Have a nice code day 
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Yip, this was my way to begin with.
regards,
Rich
"Programming today is a race between software engineers striving to build bigger and
better idiot-proof programs, and the Universe trying to produce bigger and better idiots.
So far the Universe is winning." -- Rich Cook
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Read this function, it could be another start point!
This function that I wrote last year find the bigger long, greater than the input value, that is a power of 2.
And it is quite fast!
unsigned long Next2Power(unsigned long x){<br />
unsigned long y=1, x1=x;<br />
if(x==0) return 0;<br />
while(x1!=0){<br />
x1>>=1;<br />
y<<=1;<br />
}<br />
y>>=1;<br />
if(y!=x) y<<=1;<br />
return y;<br />
}
Have a nice code day
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_Russell_ wrote: This function that I wrote last year find the bigger long, greater than the input value, that is a power of 2.
Why not just use log<sub>2</sub>()?
"Let us be thankful for the fools. But for them the rest of us could not succeed." - Mark Twain
"There is no death, only a change of worlds." - Native American Proverb
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Because my function is faster (I think) than log2().
But if I'm wrong let me know how have a faster function. And why.
thanks
Have a nice code day
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_Russell_ wrote: ...let me know how have a faster function.
This one benchmarks faster than the one you've shown:
unsigned long Next2Power(unsigned long x)
{
x--;
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x++;
return x;
}
"Let us be thankful for the fools. But for them the rest of us could not succeed." - Mark Twain
"There is no death, only a change of worlds." - Native American Proverb
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Hello RichardS,
Try this one.
double val = 0.0;
val = log10(y) / log10(2.0);
int val2 = int(val);
if (val == val2)
else
here 'y' is value u want to check for power of 2.
Best of luck
Divyang Mithaiwala
System Engineer & Software Developer
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we are on a standard C++ forum, not managed/CLI...
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Hello v2.0,
We are on visual C++ & my suggesion is not wrong for it.
Divyang Mithaiwala
System Engineer & Software Developer
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Divyang Mithaiwala wrote: my suggesion is not wrong for it.
yes it is, it will not work on a MFC/Win32 project...
it is only for managed C++ (which is not for this forum, but for the C++/CLI forum instead)
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Here logic is simple
y = value that want to check
x = find a value that is in power of 2
so,
2^x = y
log 2^x = log y
x log 2 = log y
x = log y /log 2
Divyang Mithaiwala
System Engineer & Software Developer
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Hi,
you could use ::log to check if its a power of 2
math:
y = 2^x -> x = log(y)/log(2)
so you can use:
int y = 1024;
if(y == (int)(::pow(2.0, ::floor(::log((double)y)/::log(2.0)))))
{
}
else
{
}
but his is heavy processing,
so it mith be slower that the for loop
codito ergo sum
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Define "fast".
If you're using 32-bit int's, there are only 32 possibilities so you can just check entries in a lookup table, making sure you do the most likely candidates first.
BTW, all these suggestions to use sqrt() and log() are nuts - these functions are insanely slow!
The two most common elements in the universe are Hydrogen and stupidity. - Harlan Ellison
Awasu 2.2 [^]: A free RSS/Atom feed reader with support for Code Project.
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Taka Muraoka wrote: BTW, all these suggestions to use sqrt() and log() are nuts - these functions are insanely slow!
Agreed!! It is quite simple. A single decrement and bitwise AND will do the job nicely
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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Crikey. I thought this was well known. Just do this:
if (x & (x-1))
{
}
else
{
}
sorry, got the two cases round the wrong way... 
Ryan "Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
-- modified at 6:04 Monday 6th March, 2006
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