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Hi,
I read this question somewhere and also found its answer on the web but couldnt figure it out. can anyone pls enlighten me:
Ques:
Test if an unsigned integer is a power of two without loops or recursion?
Soln: #define power_of_two(x) \ ((x)&&(~(x&(x-1))))
Can anyone figure out what is this macro doing?
Pls try to simplify ur explanation as much as possible.
Thanks.
Here's a little different code for the same thing:
Returns 1 is it is a power of two, zero otherwise. The code changes the least significant bit of x to zero, and if x is a power of two then after the statement (x & (x - 1)) there is only one bit with a value of one and if you set it to zero it leaves you with zero for the answer.Code:#define power_of_two(x) ((x & (x - 1)) == 0);
Hey Entropic,
Thanks a lot for the code and the explanation.
#define PTWO(x) (x%2==0)
Quzah.
Nono, that checks whether the number is odd or even, you little brat :pQuote:
Originally posted by quzah
#define PTWO(x) (x%2==0)
Quzah.
Yeah. I was thinking if it was a multiple of two, not a power of...Quote:
Originally posted by Magos
Nono, that checks whether the number is odd or even, you little brat :p
Quzah.
Since the quick way has been done for you, I'll give you a different way:
Unlike the first example, mine will work on floating point numbers. Neither example will work on negative numbers.Code:#define pot_test(x) \
( \
x==(1<<1) || \
x==(1<<2) || \
x==(1<<3) || \
x==(1<<4) || \
x==(1<<5) || \
x==(1<<6) || \
x==(1<<7) || \
x==(1<<8) || \
x==(1<<9) || \
x==(1<<10) || \
x==(1<<11) || \
x==(1<<12) || \
x==(1<<13) || \
x==(1<<14) || \
x==(1<<15) || \
x==(1<<16) || \
x==(1<<17) || \
x==(1<<18) || \
x==(1<<19) || \
x==(1<<20) || \
x==(1<<21) || \
x==(1<<22) || \
x==(1<<23) || \
x==(1<<24) || \
x==(1<<25) || \
x==(1<<26) || \
x==(1<<27) || \
x==(1<<28) || \
x==(1<<29) || \
x==(1<<30) || \
x==(1<<31) \
)
Quzah.
You forgot 1. 2^0 = 1.
I actually intentionally left it out.Quote:
Originally posted by Magos
You forgot 1. 2^0 = 1.
Quzah.