Coarray SYNC ALL issue

OP1 · ‎10-26-2016

The following code exhibits a different behavior depending on whether line 3 or 4 is commented out. It seems that the SYNC ALL behavior varies depending on which image 'finishes first'.

PROGRAM MAIN
IMPLICIT NONE
IF (THIS_IMAGE()==1) THEN   ! This works.
!IF (THIS_IMAGE()==2) THEN   ! Any image number > 1 causes the code to hang at the SYNC ALL below.
    CALL SLEEP(1)
    WRITE(*,*) 'Image ',THIS_IMAGE(),' is done!'
    GO TO 100
END IF
SYNC ALL
WRITE(*,*) 'Image ',THIS_IMAGE(),' is done!'
100 CONTINUE
END PROGRAM MAIN

Steven_L_Intel1 · ‎10-26-2016

This program is malformed - a SYNC ALL waits for all other images to execute a SYNC ALL the same number of times. If you skip over the SYNC ALL and let the program exit, the behavior is unspecified. What probably happens in the case image 1 skips it is that this image is "special" and it exiting causes all other images to exit.

OP1 · ‎10-27-2016

Thanks Steve for the clarification. I think I got confused with the fact that when an image stops, its coarrays are still accessible (see code below). I thought that therefore a stopped image would not be considered in a SYNC ALL statement (only the images still running would wait for each other).

PROGRAM MAIN
IMPLICIT NONE
INTEGER :: I,J
IF (THIS_IMAGE()==1) THEN
    CALL SLEEP(1)
    DO I=1,NUM_IMAGES()
        J = I
    END DO
    SYNC MEMORY
    DO I=1,NUM_IMAGES()
        WRITE(*,*) J
    END DO
END IF
WRITE(*,*) 'Image ',THIS_IMAGE(),' is done!'
END PROGRAM MAIN

Steven_L_Intel1 · ‎10-27-2016

Fortran 2008 doesn't take into account stopped or failed images - this is something being added for Fortran 2015. In F2015, if there are stopped or failed images in the current team (another F2015 thing), a SYNC ALL without a STAT= is an error and will cause termination.

In Fortran 2008, it is assumed that all images are still active.