A Cache Efficient Matrix Transpose Program?

Question

So the obvious way to transpose a matrix is to use :

  for( int i = 0; i < n; i++ )

    for( int j = 0; j < n; j++ )

      destination[j+i*n] = sourc         


        

Answer 1

I had the exact same problem yesterday. I ended up with this solution:

void transpose(double *dst, const double *src, size_t n, size_t p) noexcept {
    THROWS();
    size_t block = 32;
    for (size_t i = 0; i < n; i += block) {
        for(size_t j = 0; j < p; ++j) {
            for(size_t b = 0; b < block && i + b < n; ++b) {
                dst[j*n + i + b] = src[(i + b)*p + j];
            }
        }
    }
}

This is 4 time faster than the obvious solution on my machine.

This solution takes care of a rectangular matrix with dimensions which are not a multiple of the block size.

if dst and src are the same square matrix an in place function should really be used instead:

void transpose(double*m,size_t n)noexcept{
    size_t block=0,size=8;
    for(block=0;block+size-1

I used C++11 but this could be easily translated in other languages.