C++ for Scientists - Technische Universität Dresden

166 CHAPTER 5. META-PROGRAMMING
}
vector u(s), v(s), w(s);
vector u(s), v(s), w(s);
for (unsigned i= 0; i < s; i++) {
v[i]= float(i);
w[i]= float(2∗i + 15);
}
for (unsigned j= 0; j < 3; j++)
for (unsigned i= 0; i < s; i++)
u[i]= 3.0f ∗ v[i] + w[i];
const unsigned rep= 200000;
boost::timer native;
for (unsigned j= 0; j < rep; j++)
for (unsigned i= 0; i < s; i++)
u[i]= 3.0f ∗ v[i] + w[i];
std::cout ≪ ”Compute time native loop is ” ≪ 1000000.0 ∗ native.elapsed() / double(rep) ≪ ” µs.\n”;
return 0 ;
Alternatively we compute this with an unrolling of 4 cycles:
for (unsigned j= 0; j < rep; j++)
for (unsigned i= 0; i < s; i+= 4) {
u[i]= 3.0f ∗ v[i] + w[i];
u[i+1]= 3.0f ∗ v[i+1] + w[i+1];
u[i+2]= 3.0f ∗ v[i+2] + w[i+2];
u[i+3]= 3.0f ∗ v[i+3] + w[i+3];
}
This code will obviously only work if the vector size is divisible by 4. To avoid errors we can
add an assertion on the vector size but this is not really satisfying. Instead, we generalize this
implementation to arbitrary vector sizes:
boost::timer unrolled;
for (unsigned j= 0; j < rep; j++) {
unsigned sb= s / 4 ∗ 4;
for (unsigned i= 0; i < sb; i+= 4) {
u[i]= 3.0f ∗ v[i] + w[i];
u[i+1]= 3.0f ∗ v[i+1] + w[i+1];
u[i+2]= 3.0f ∗ v[i+2] + w[i+2];
u[i+3]= 3.0f ∗ v[i+3] + w[i+3];
}
for (unsigned i= sb; i < s; i++)
u[i]= 3.0f ∗ v[i] + w[i];
}
std::cout ≪ ”Compute time unrolled loop is ” ≪ 1000000.0 ∗ unrolled.elapsed() / double(rep) ≪ ” µs.\n”;
std::cout ≪ ”u is ” ≪ u ≪ ’\n’;
Listing 5.4: Unrolled computation of u = 3v + w
The little program was compiled with g++ 4.1.2 with the flags -O3 -ffast-math -DNDEBUG