arguments - What is the explicit difference between the fortran intents (in,out,inout)?

Question

After searching for a while in books, here on stackoverflow and on the general web, I have found that it is difficult to find a straightforward explanation to the real differences between the fortran argument intents. The way I have understood it, is this:

• intent(in) -- The actual argument is copied to the dummy argument at entry.
• intent(out) -- The dummy argument points to the actual argument (they both point to the same place in memory).
• intent(inout) -- the dummy argument is created locally, and then copied to the actual argument when the procedure is finished.

If my understanding is correct, then I also want to know why one ever wants to use intent(out), since the intent(inout) requires less work (no copying of data).

Answer 1

Intents are just hints for the compiler, and you can throw that information away and violate it. Intents exists almost entirely to make sure that you only do what you planned to do in a subroutine. A compiler might choose to trust you and optimize something.

This means that intent(in) is not pass by value. You can still overwrite the original value.

program xxxx  
    integer i  
    i = 9  
    call sub(i)  
    print*,i ! will print 7 on all compilers I checked  
end  
subroutine sub(i)  
    integer,intent(in) :: i  
    call sub2(i)  
end  
subroutine sub2(i)  
    implicit none  
    integer i  
    i = 7  ! This works since the "intent" information was lost.  
end

program xxxx  
    integer i  
    i = 9  
    call sub(i)  
end  
subroutine sub(i)  
    integer,intent(out) :: i  
    call sub2(i)  
end  
subroutine sub2(i)  
    implicit none   
    integer i  
    print*,i ! will print 9 on all compilers I checked, even though intent was "out" above.  
end