C / C++ / MFC

double val = 0.0;
val = log10(y) / log10(2.0);

//Check that 'val' is pure int (Here you use ur own code is u optimize)
int val2 = int(val);

if (val == val2) 
     // y is power of 2
else
     // y is not power of 2

Divyang Mithaiwala wrote:

my suggesion is not wrong for it.

2^x = y
log 2^x = log y
x log 2 = log y

x = log y /log 2 

int y = 1024;
if(y == (int)(::pow(2.0, ::floor(::log((double)y)/::log(2.0)))))
{
  //  true
}
else
{
  // false
}

Taka Muraoka wrote:

BTW, all these suggestions to use sqrt() and log() are nuts - these functions are insanely slow!

if (x & (x-1))
{
    // x is <edit>not</edit>a power of two
}
else
{
    // x is <edit>not</edit> a power of two
}

sorry, got the two cases round the wrong way...

RichardS wrote:

Is there is quick way for now making x a power of 2 after you have decided that it is not?

#define  CK_ISPOW2(n)        ( ((n)&((n)-1)) == 0 )
 
ullong  FkGrainDn2( ullong S, ullong G )
{
	// if( !CK_ISPOW2(G) )  return( FkGrainDnN(S, G) );
	S = S & ~(G-1);
	return(S);
}
 
ullong  FkGrainUp2( ullong S, ullong G )
{
	// if( !CK_ISPOW2(G) )  return( FkGrainUpN(S, G) );
	G -= 1;
	S  = (S+G) & ~G;
	return(S);
}

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