How to write log base 2 in c c

Question

Is there any way to write log base 2  function   The  C language has 2 built in function --    1 log which is base e   2 log10 base 10   But I need log function of base 2 How to calculate this

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log2 int n    31 -   builtin clz n

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Simple math           log2  x    logy  x    logy  2   where y can be anything  which for standard log functions is either 10 or e

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uint16 t log2 uint32 t n     but truncated      if  n  0  throw          uint16 t logValue   -1       while  n               logValue             n  gt  gt   1              return logValue       Basically the same as tomlogic s

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If you re looking for an integral result  you can just determine the highest bit set in the value and return its position

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As stated on http   en wikipedia org wiki Logarithm   logb x    logk x    logk b    Which means that   log2 x    log10 x    log10 2

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I needed to have more precision that just the position of the most significant bit  and the microcontroller I was using had no math library  I found that just using a linear approximation between 2 n values for positive integer value arguments worked well   Here is the code   uint16 t approx log base 2 N times 256 uint16 t n        uint16 t msb only   0x8000      uint16 t exp   15       if  n    0          return  -1       while   n  amp  msb only     0            msb only  gt  gt   1          exp--             return    uint16 t     uint32 t   n   msb only    lt  lt  8    msb only      exp  lt  lt  8        In my main program  I needed to calculate N   log2 N    2 with an integer result   temp      uint32 t  N    approx log base 2 N times 256    512   and all 16 bit values were never off by more than 2

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C99 has log2  as well as log2f and log2l for float and long double

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If you want to make it fast  you could use a lookup table like in Bit Twiddling Hacks  integer log2 only    uint32 t v     find the log base 2 of 32-bit v int r          result goes here  static const int MultiplyDeBruijnBitPosition 32         0  9  1  10  13  21  2  29  11  14  16  18  22  25  3  30    8  12  20  28  15  17  24  7  19  27  23  6  26  5  4  31     v    v  gt  gt  1     first round down to one less than a power of 2  v    v  gt  gt  2  v    v  gt  gt  4  v    v  gt  gt  8  v    v  gt  gt  16   r   MultiplyDeBruijnBitPosition  uint32 t  v   0x07C4ACDDU   gt  gt  27     In addition you should take a look at your compilers builtin methods like  BitScanReverse which could be faster because it may entirely computed in hardware   Take also a look at possible duplicate How to do an integer log2   in C

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All the above answers are correct  This answer of mine below can be helpful if someone needs it  I have seen this requirement in many questions which we are solving using C   log2  x    logy  x    logy  2   However  if you are using C language and you want the result in integer  you can use the following   int result    int  floor log x    log 2      1    Hope this helps

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define M LOG2E 1 44269504088896340736    log2 e   inline long double log2 const long double x       return log x    M LOG2E       multiplication may be faster than division

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You have to include math h  C  or cmath  C    Of course keep on mind that you have to follow the maths that we know   only numbers 0   Example     include  lt iostream gt   include  lt cmath gt  using namespace std   int main        cout lt  lt log2 number

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Consult your basic mathematics course  log n   log 2  It doesn t matter whether you choose log or log10in this case  dividing by the log of the new base does the trick

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Improved version of what Ustaman Sangat did  static inline uint64 t log2 uint64 t n        uint64 t val      for  val   0  n  gt  1  val    n  gt  gt   1        return val

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log2 x    log10 x    log10 2

[c++] How to write log base(2) in c/c++

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