What is the most efficient way to enumerate vertices of k dimensional hypercube in C++?

Basic Question: I have a k dimensional box. I have a vector of upper bounds and lower bounds. What is the most efficient way to enumerate the coordinates of the vertices?

Background: As an example, say I have a 3 dimensional box. What is the most efficient algorithm / code to obtain:

vertex[0] = ( 0, 0, 0 ) -> ( L_0, L_1, L_2 )
vertex[1] = ( 0, 0, 1 ) -> ( L_0, L_1, U_2 )
vertex[2] = ( 0, 1, 0 ) -> ( L_0, U_1, L_2 )
vertex[3] = ( 0, 1, 1 ) -> ( L_0, U_1, U_2 )

vertex[4] = ( 1, 0, 0 ) -> ( U_0, L_1, L_2 )
vertex[5] = ( 1, 0, 1 ) -> ( U_0, L_1, U_2 )
vertex[6] = ( 1, 1, 0 ) -> ( U_0, U_1, L_2 )
vertex[7] = ( 1, 1, 1 ) -> ( U_0, U_1, U_2 )

where L_0 corresponds to the 0'th element of the lower bound vector & likewise U_2 is the 2nd element of the upper bound vector.

My Code:

const unsigned int nVertices = ((unsigned int)(floor(std::pow( 2.0, double(nDimensions)))));

for ( uns开发者_StackOverflow社区igned int idx=0; idx < nVertices; ++idx )
{
   for ( unsigned int k=0; k < nDimensions; ++k )
   {
      if ( 0x00000001 & (idx >> k) )
      {
         bound[idx][k] = upperBound[k];
      }
      else
      {
         bound[idx][k] = lowerBound[k];
      }
   }
}

where the variable bound is declared as:

std::vector< std::vector<double> > bound(nVertices);

but I've pre-sized it so as not to waste time in the loop allocating memory. I need to call the above procedure about 50,000,000 times every time I run my algorithm -- so I need this to be really efficient.

Possible Sub-Questions: Does it tend to be faster to shift by k instead of always shifting by 1 and storing an intermediate result? (Should I be using >>= ??)

It will probably go faster if you can reduce conditional branching:

bound[idx][k] = upperLowerBounds[(idx >> k) & 1][k];

You might improve things even more if you can interleave the upper and lower bounds in a single array:

bound[idx][k] = upperLowerBounds[(k << 1) | (idx >> k)&1];

I don't know if shifting idx incrementally helps. It's simple enough to implement, so it's worth a try.