What is the fastest way to calculate the number of bits needed to store a number

问题

I'm trying to optimize some bit packing and unpacking routines. In order to do the packing I need to calculate the number of bits needed to store integer values. Here is the current code.

if (n == -1) return 32;
if (n == 0) return 1;
int r = 0;
while (n)
{
    ++r;
    n >>= 1;
}
return r;

回答1:

You're looking to determine the integer log base 2 of a number (the l=highest bit set). Sean Anderson's "Bit Twiddling Hacks" page has several methods ranging from the obvious counting bits in a loop to versions that use table lookup. Note that most of the methods demonstrated will need to be modified a bit to work with 64-bit ints if that kind of portability is important to you.

http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogObvious

Just make sure that any shifting you're using to work out the highest bit set needs to be done' on an unsigned version of the number since a compiler implementation might or might not sign extend the >> operation on a signed value.

回答2:

Non-portably, use the bit-scan-reverse opcode available on most modern architectures. It's exposed as an intrinsic in Visual C++.

Portably, the code in the question doesn't need the edge-case handling. Why do you require one bit for storing 0? In any case, I'll ignore the edges of the problem. The guts can be done efficiently thus:

if (n >> 16) { r += 16; n >>= 16; }
if (n >>  8) { r +=  8; n >>=  8; }
if (n >>  4) { r +=  4; n >>=  4; }
if (n >>  2) { r +=  2; n >>=  2; }
if (n - 1) ++r;

回答3:

What you are trying to do is find the most significant bit. Some architectures have a special instruction just for this purpose. For those that don't, use a table lookup method.

Create a table of 256 entries, wherein each element identifies the upper most bit.

Either loop through each byte in the number, or use a few if-statements to break to find the highest order non-zero byte.

I'll let you take the rest from here.

回答4:

Do a binary search instead of a linear search.

if ((n >> 16) != 0)
{
    r += 16;
    n >>= 16;
}

if ((n >> 8) != 0)
{
    r += 8;
    n >>= 8;        
}

if ((n >> 4) != 0)
{
    r += 4;
    n >>= 4;        
}

// etc.

If your hardware has bit-scan-reverse, an even faster approach would be to write your routine in assembly language. To keep your code portable, you could do

#ifdef ARCHITECTURE_WITH_BSR
   asm // ...
#else
   // Use the approach shown above
#endif

回答5:

You would have to check the execution time to figure the granularity, but my guess is that doing 4 bits at a time, and then reverting to one bit at a time would make it faster. Log operations would probably be slower than logical/bit operations.

if (n < 0) return 32;
int r = 0;
while (n && 0x7FFFFFF0) {
  r+=4;
  n >>= 4; }
while (n) {
  r++;
  n >>= 1; }
return r;

回答6:

number_of_bits = log2(integer_number)

rounded to the higher integer.

来源：https://stackoverflow.com/questions/2721244/what-is-the-fastest-way-to-calculate-the-number-of-bits-needed-to-store-a-number

标签

c++

bit-manipulation

bits