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Simplification of FINDLOC with array of size 0 fails

Harald1
New Contributor II
1,110 Views

The following code is rejected by ifort 2021.6.0:

program p
! integer, parameter :: ii(1) = findloc ([integer::1],1) ! works
  integer, parameter :: ii(1) = findloc ([integer:: ],1) ! fails
  print *, ii
  print *, findloc ([integer::],1) ! accepted
end

This gives:

% ifort ifort-findloc.f90
ifort-findloc.f90(3): error #6821: A scalar constant is required in this context.
  integer, parameter :: ii(1) = findloc ([integer:: ],1) ! fails
--------------------------------^
compilation aborted for ifort-findloc.f90 (code 1)

It shouldn't be too hard to do the simplification...

 

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Barbara_P_Intel
Employee
1,015 Views

I filed a bug report on this erroneous error message, CMPLRIL0-34931.



View solution in original post

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Mark_Lewy
Valued Contributor I
1,051 Views

I'd argue that this is a compiler bug as

findloc ([integer:: ],1)

looks like a legitimate constant expression (according to MFE 8.4).

I note also that gfortran --warn-all (MSYS2 10.2.0) compiles this without any warnings and produces the desired output.

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Barbara_P_Intel
Employee
1,016 Views

I filed a bug report on this erroneous error message, CMPLRIL0-34931.



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Barbara_P_Intel
Employee
810 Views

This FINDLOC() issue is resolved in the latest Fortran compilers that were released in December. Give it a try!



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Harald1
New Contributor II
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Yes, it is fixed in the latest version.  Thanks!

 

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