Concatenate boost::dynamic_bitset or std::bitset

Question

what is the best way to concatenate 2 bitsets?

For example i\'ve got

boost::dynamic_bitset<> test1( std::string(\"1111\") );
boost::dynamic_bit         


        

Answer 1

I ran a test comparing the following two approaches:

  /* ... */
  for( int ii = 0; ii < 1000000; ++ii ) {
    std::bitset<16> v1( randomUlongs[ii] );
    std::bitset<16> v2( randomUlongs[ii+1] );
#ifdef STRING_TEST
    std::bitset<32> v3( v1.to_string() + v2.to_string() );
#else
    std::bitset<32> v3( v2.to_ulong() | (v1.to_ulong() << 16) );
    /* print out v3 */
  }

... where randomUlongs was constant during each run (a large array in a header) to avoid anything contaminating results. I timed it with:

~ time for ((ii=0; ii<1000; ii++)); do ./bitset_test >/dev/null; done

Under Linux (x86_i686) with gcc 4.4.6 at optimization level 3: the string concatenation was fastest, by a factor of 2.

Under Solaris (sparc) with gcc 3.4.3 and Sun Studio C++ 5.12 (2011/11/16), both with optimization level 3: the non-string approach was fastest by a factor of 10.

I think you'll find the "fastest" solution is highly dependent on the compiler, though I suppose platform could play a significant role as well.