How do I ..?

I would make the test like this:

float fPower = log10(nInput) / log10(2);
float fRound = (float)floor(fPower+0.5f);
if( (float)fabs(fPower - fRound) < 1e-5f )
{
// The number is power of two
}

You could also count the number of set bits in nInput. If there are only one bit set then you know nInput is power of two. If this is faster or not I don''t know.

- WitchLord

My funcion would look like this.

bool ispower2(int num)
{
float exponet = log10(num) / log10(2);
int testnum;

testnum = (int)exponet;
if(exponet == testnum) {
// then its a power of two
return true;
}

return false;
}

hope that helps


Edited by - UraniumRod on 4/29/00 6:29:33 PM

on a side note, you can do this:

return exponet == testnum;

instead of the if, else block


adamm@san.rr.com


bool IsPowerOf2(unsigned int nInput)
{
unsigned char bitcounter = 0;
for(unsigned char i=0;(i<32) && (bitcounter<2);i++)
{
bitcounter += nInput&1;
nInput >>= 1;
}
return (bitcounter==1);
}


If you want to test if an integer is a power of 2, you don't need to work with floating point operations.

Visit our homepage: www.rarebyte.de.st

GA

Edited by - ga on 5/2/00 3:13:31 PM

Could you explain me that code? I think I''m gonna read me a good C++ book and finish it for a change ;-)..

----------
Drago, VerSoft