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Just a simple of boolean true/false that is very efficient would be nice. Should I use recursion, or is there some better way to determine it?
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Another solution is to keep dividing the number by two, i.e, do n = n/2 iteratively. In any iteration, if n%2 becomes non-zero and n is not 1 then n is not a power of 2. If n becomes 1 then it is a power of 2.
A simple method for this is to simply take the log of the number on base 2 and if you get an integer then the number is the power of 2.
Algorithm to Check IF a Number is Power of Another NumberFind the log of a base b and assign its integer part to variable x. Also, find the b to the power x and assign it to another variable y. Check if y is equal to a then a is a power of another number b and print a is the power of another number b.
From here:
Determining if an integer is a power of 2
unsigned int v; // we want to see if v is a power of 2 bool f; // the result goes here f = (v & (v - 1)) == 0;Note that 0 is incorrectly considered a power of 2 here. To remedy this, use:
f = v && !(v & (v - 1));
Why does this work? An integer power of two only ever has a single bit set. Subtracting 1 has the effect of changing that bit to a zero and all the bits below it to one. AND'ing that with the original number will always result in all zeros.
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