Unable to print allocated status of a allocatable inside a derived type

Question

I would like to know why this code returns error in the last print.

With gfortran 7.4.0 fails but with ifort 18.0.3 works well.

program test
implicit none
type :: syntax
  integer, allocatable :: f(:)
end type
type(syntax), allocatable :: rhs(:)

allocate(rhs(2))
print*, allocated(rhs(2)%f)
print*, allocated(rhs(size(rhs))%f)
end program

The gfortran error is:

 F

Program received signal SIGSEGV: Segmentation fault - invalid memory reference.

Backtrace for this error:
#0  0x7f4cf40442da in ???
#1  0x7f4cf4043503 in ???
#2  0x7f4cf3c76f1f in ???
#3  0x55aa522e5e50 in test
    at /home/pena/Escritorio/c.f90:10
#4  0x55aa522e5f0d in main
    at /home/pena/Escritorio/c.f90:11
Violación de segmento (`core' generado)

What is the result with ifort? What is printed when you place between the 2 print statements `print *, size(rhs)`. With gfortran 7.3.0 (Cygwin) I get `F` `2` `T`. — albert, Nov 06 '19 at 15:44

score 4 · Accepted Answer · answered Nov 06 '19 at 15:57

4

This is a bug in gfortran which is not present in version 8.

If you can't upgrade your compiler, then there is an easy alternative: simply use a temporary variable for size(rhs):

hack = SIZE(rhs)
print*, allocated(rhs(hack)%f)

giving output with gfortran 7.4.0:

 F
 F

answered Nov 06 '19 at 15:57

francescalus

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Thank you, I will do it. – franpena Nov 06 '19 at 15:58

Unable to print allocated status of a allocatable inside a derived type

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