Reputation: 31
I tried to rewrite code from Fortran to C++ with a 2000*2000 matrix multiplication implements through Eigen library. I found that for loop in Eigen is much slower (>3x) than do loop in Fortran. The codes are listed below:
test.f90
program main
implicit none
integer :: n,i,j,k
integer :: tic,toc
real(8),ALLOCATABLE ::a(:,:),b(:,:),c(:,:)
real(8) :: s
n = 2000
allocate(a(n,n),b(n,n),c(n,n))
do i=1,n
do j =1,n
a(j,i) = i * 1.0
b(j,i) = i * 1.0
enddo
enddo
call system_clock(tic)
do j=1,n
do i=1,n
s = 0.0
do k=1,n
s = s + a(i,k) * b(k,j)
enddo
c(i,j) = s
enddo
enddo
call system_clock(toc)
print*,'Fortran with loop:', (toc - tic) / 1000.0
call system_clock(tic)
c = matmul(a,b)
call system_clock(toc)
print*,'Fortran with matmul:', (toc - tic) / 1000.0
DEALLOCATE(a,b,c)
end
test.cpp
#include<Eigen/Core>
#include<time.h>
#include<iostream>
using Eigen::MatrixXd;
int main(){
int n = 2000;
MatrixXd a(n,n),b(n,n),c(n,n);
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
a(i,j) = i * 1.0;
b(i,j) = j * 1.0;
}
}
clock_t tic,toc;
tic = clock();
for(int j=0;j<n;j++){
for(int i=0;i<n;i++){
double s= 0.0;
for(int k=0;k<n;k++){
s += a(i,k) * b(k,j);
}
c(i,j) = s;
}
}
toc = clock();
std::cout << (double)((toc - tic)) / CLOCKS_PER_SEC << std::endl;
tic = clock();
c= a * b;
toc = clock();
std::cout << (double)((toc - tic)) / CLOCKS_PER_SEC << std::endl;
}
Compiled by(with gcc-8.4, in Ubuntu-18.04)
gfortran test.f90 -O3 -march=native -o testf
g++ test.cpp -O3 -march=native -I/path/to/eigen -o testcpp
And I get results:
Fortran with loop: 10.9700003
Fortran with matmul: 0.834999979
Eigen with loop: 38.2188
Eigen with *: 0.40625
The internal implementation is of comparable speed, but why Eigen is much slower for the loop implementation?
Upvotes: 3
Views: 1059
Reputation: 749
The biggest problem with the loops is that they are not done in the proper order for either C++ (which should be row-major), or Fortran (which should be column-major). This gives you a large performance hit, especially for large matrices.
The nativemul
implementation by John Alexiou (with dot_product
) has the same problem, so I am very surprised that he claims it's faster. (And I find that it isn't; see below. Maybe his (intel?) compiler rewrites the code to use matmul internally.)
This is the correct loop order for Fortran:
c = 0
do j=1,n
do k=1,n
do i=1,n
c(i,j) = c(i,j) + a(i,k) * b(k,j)
enddo
enddo
enddo
With gfortran version 10.2.0, and compiled with -O3, I get
Fortran with original OP's loop: 53.5190010
Fortran with John Alexiou's nativemul: 53.4309998
Fortran with correct loop: 11.0679998
Fortran with matmul: 2.3699999
A correct loop in C++ should give you similar performance.
Obviously matmul/BLAS are much faster for large matrices.
Upvotes: 3
Reputation: 29244
In the Fortran code I saw the same problem, but then I moved the matrix multiplication in a subroutine and the resultant speed was almost as good as matmul
. I also compared to BLAS Level 3 function.
Fortran with loop: 9.220000
Fortran with matmul: 8.450000
Fortran with blas3: 2.050000
and the code to produce it
program ConsoleMatMul
use BLAS95
implicit none
integer :: n,i,j
integer :: tic,toc
real(8),ALLOCATABLE :: a(:,:),b(:,:),c(:,:),xe(:,:)
n = 2000
allocate(a(n,n),b(n,n),c(n,n),xe(n,n))
do i=1,n
do j =1,n
a(j,i) = i * 1.0
b(j,i) = i * 1.0
enddo
enddo
call system_clock(tic)
call nativemul(a,b,c)
call system_clock(toc)
print*,'Fortran with loop:', (toc - tic) / 1000.0
call system_clock(tic)
c = matmul(a,b)
call system_clock(toc)
print*,'Fortran with matmul:', (toc - tic) / 1000.0
c = b
xe = 0d0
call system_clock(tic)
call gemm(a,c,xe) ! BLAS MATRIX/MATRIX MUL
call system_clock(toc)
print*,'Fortran with blas3:', (toc - tic) / 1000.0
DEALLOCATE(a,b,c)
contains
pure subroutine nativemul(a,b,c)
real(8), intent(in), allocatable :: a(:,:), b(:,:)
real(8), intent(out), allocatable :: c(:,:)
real(8) :: s
integer :: n, i,j,k
n = size(a,1)
if (.not. allocated(c)) allocate(c(n,n))
do j=1,n
do i=1,n
s = 0.0d0
do k=1,n
s = s + a(i,k) * b(k,j)
end do
c(i,j) = s
end do
end do
end subroutine
end program ConsoleMatMul
before I moved the code into a subroutine I got
Fortran with loop: 85.450000
Update the native multiplication reaches matmul
levels (or exceeds it) when the inner loop is replaced by a dot_product()
.
pure subroutine nativemul(a,b,c)
real(8), intent(in) :: a(:,:), b(:,:)
real(8), intent(out) :: c(:,:)
integer :: n, i,j
n = size(a,1)
do j=1,n
do i=1,n
c(i,j) = dot_product(a(i,:),b(:,j))
! or = sum(a(i,:)*b(:,j))
end do
end do
end subroutine
Upvotes: 0
Reputation: 74
C++ pre-increment is faster than post-increment...
for(int j=0;j<n;++j){
for(int i=0;i<n;++i){
double s= 0.0;
for(int k=0;k<n;++k){
s += a(i,k) * b(k,j);
}
c(i,j) = s;
}
}
Upvotes: -1