gcc/testsuite/gfortran.dg/matmul_3.f90

   1 ! { dg-do run }
   2 ! Check the fix for PR28005, in which the mechanism for dealing
   3 ! with matmul (transpose (a), b) would cause wrong results for
   4 ! matmul (a(i, 1:n), b(1:n, 1:n)).
   5 !
   6 ! Based on the original testcase contributed by
   7 ! Tobias Burnus  <tobias.burnus@physik.fu-berlin.de>
   8 !
   9    implicit none
  10    integer, parameter         ::  nmax = 3
  11    integer                    ::  i, n = 2
  12    integer, dimension(nmax,nmax) ::  iB=0 , iC=1
  13    integer, dimension(nmax,nmax) ::  iX1=99, iX2=99, iChk
  14    iChk = reshape((/30,66,102,36,81,126,42,96,150/),(/3,3/))
  15
  16 ! This would give 3, 3, 99
  17    iB = reshape((/1 ,3 ,0 ,2 ,5 ,0 ,0 ,0 ,0 /),(/3,3/))
  18    iX1(1:n,1) = matmul( iB(2,1:n),iC(1:n,1:n) )
  19
  20 ! This would give 4, 4, 99
  21    ib(3,1) = 1
  22    iX2(1:n,1) = matmul( iB(2,1:n),iC(1:n,1:n) )
  23
  24 ! Whereas, we should have 8, 8, 99
  25    if (any (iX1(1:n+1,1) .ne. (/8, 8, 99/))) STOP 1
  26    if (any (iX1 .ne. iX2)) STOP 2
  27
  28 ! Make sure that the fix does not break transpose temporaries.
  29    iB = reshape((/(i, i = 1, 9)/),(/3,3/))
  30    ic = transpose (iB)
  31    iX1 = transpose (iB)
  32    iX1 = matmul (iX1, iC)
  33    iX2 = matmul (transpose (iB), iC)
  34    if (any (iX1 .ne. iX2)) STOP 3
  35    if (any (iX1 .ne. iChk)) STOP 4
  36 end