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Just out of curiosity, how can you tell if a number x is a power of two (x = 2^n) without using recursion.
Thanks
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A simple solution is to repeatedly compute powers of x. If a power becomes equal to y, then y is a power, else not.
pow() is function to get the power of a number, but we have to use #include<math. h> in c/c++ to use that pow() function. then two numbers are passed. Example – pow(4 , 2); Then we will get the result as 4^2, which is 16.
Number is a power of 10 if it's equal to 10, 100, 1000 etc. 1 is also 0-th power of 10. Other numbers like 2, 3, 11, 12 etc. are not powers of 10.
One way is to use bitwise AND. If a number $x is a power of two (e.g., 8=1000), it will have no bits in common with its predecessor (7=0111). So you can write:
($x & ($x - 1)) == 0 Note: This will give a false positive for $x == 0.
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