Powers of 2
Author
Is there a simple way to find out if an (int) number is a power of 2?
I know only one bit can be 1 and the last bit should be 0, but how do I find out in an easy function?
Marty
const bool isPowerOf2(int num){
return (num!=0) && ((num&(num-1))==0);
}
emptyhead
Watch out for advice from the successfull, they don''t want company
return (num!=0) && ((num&(num-1))==0);
}
emptyhead
Watch out for advice from the successfull, they don''t want company
// checks if a number is a power of 2
BOOL IsPow2(int i)
{
// & the negative 2''s complement number with the passed param
// a power of 2 number will slip through
if ( (i & -i) == i ) return TRUE;
else return FALSE;
}// END IsPow2
There you go
BOOL IsPow2(int i)
{
// & the negative 2''s complement number with the passed param
// a power of 2 number will slip through
if ( (i & -i) == i ) return TRUE;
else return FALSE;
}// END IsPow2
There you go
heres another way (the last one was good though):
given 2^y= z, if z is you number and it is a power of 2, y must be an int, or if y = lnz/ln2 (ln = log, you can use log10 - it doesn't matter). Note: z must be greater than 0.
you can also edit this funciton to return which power of two the number is, like so:
[edited by - llvllatrix on March 2, 2003 11:49:29 AM]
given 2^y= z, if z is you number and it is a power of 2, y must be an int, or if y = lnz/ln2 (ln = log, you can use log10 - it doesn't matter). Note: z must be greater than 0.
#include <math.h>#define LN_2 0.69314718bool Power2(int value){ if(value <= 0) reutrn false; if( (log((double)value)/LN_2)%1 == 0) return true; else return false;} you can also edit this funciton to return which power of two the number is, like so:
#include <math.h>#define LN_2 0.69314718int Power2(int value){ if(value <= 0) reutrn 0; float x; x = float(log((double)value)/LN_2); if( x%1 == 0) return (int)x; else return 0;} [edited by - llvllatrix on March 2, 2003 11:49:29 AM]
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