What happens when a subroutine expecting an integer*4 argument is called with a integer*8 argument?
The answer depends on the hardware and software platform, and unfortunately errors can often be difficult to detect. To understand this, look at the memory layout of the number 12345 stored in integer*4 and integer*8 variables, on an regular Intel CPU (x86):
39 30 0 0
39 30 0 0 0 0 0 0
The first line shows the four bytes of the integer*4, in the order they are stored in memory on this particular machine. The second line shows the eight bytes of the integer*8 representing the same number. As you can see the start of the integer*8 is the same as the integer*4 (on this machine!).
Similarly, the number -1 is stored as
ff ff ff ff
ff ff ff ff ff ff ff ff
Again the first four bytes are the same.
To understand what can go wrong take the FORTRAN77 example
subroutine f(x)
integer*4 x
print *, x
x = 12345
end
program p
integer*8 y
y = -1
call f(y)
print *,y
end
Since Fortran arguments are passed by reference, the subroutine f will get a pointer to the variable y. From the example above we know that the first four bytes of an integer*8 “-1” are the same as those of a integer*4 “-1”.
Therefore the number “-1” will be printed from f(), after which things go bad. This is what happens:
First we set y = -1:
y = ff ff ff ff ff ff ff ff
Then f(y) is called. The variable x will reference “y”, but will only use the first four bytes:
x = ff ff ff ff (ff ff ff ff)
This x is printed as “-1”. The we set x = 12345, note that only the first four bytes are modified!
x = 39 30 00 00 (ff ff ff ff)
When f() returns y will have the value
y = 39 30 00 00 ff ff ff ff
Which will be printed as -4294954951.
This is just one example of how things can go wrong, and why things can often work by “chance”.
Note that this illustrated situation is very platform dependent; on a big-endian CPU the above example would not work at all. Always make sure to use the correct integer type in arguments, and link to correct versions of external libraries.
FORTRAN90 compilers, which are checking also agreement of data types, should catch most of their incompatibilities in the code.
Though this is not the case of the Fortran77 example given above, one can rewrite this small piece of the code into a proper ‘Fortran90 checking form’ by utilizing module.
Program “test_f90.F90” (don’t forget big F in the suffix when applying preprocessor statements!) shall have this form:
module pass_integer
contains
subroutine f(x)
#if defined (F_INT4)
integer(kind=4), intent(inout):: x! F90 compiler gives error!
#else
integer(kind=8), intent(inout):: x! F90 compiler passed
#endif
print *, x
x = 12345
end subroutine f
end module pass_integer
program p
use pass_integer
integer (kind=8):: y
y = -1
call f(y)
print *,y
end program p
Typing
gfortran -DF_INT4 test_f90.F90
gives error of “Type/rank mismatch in argument ‘x’” while
gfortran test_f90.F90
compiles with the proper integer kind.