Debanjan Basu - 2 months ago 12

Python Question

Experience:

fortran for about 3 months

python - intermediate : never used the ctypes module in python before this

I was looking for a way to use the fortran code for my doctoral work in python - subsequently using the computation for visualizations on-the-fly for visualizations using matplotlib.

THIS POST helped (this tells that fortran code can be used/invoked in python using the ctypes module - and given that the fortran functions have alternate names bound to them - this much makes sense to me logically although i do not know how this works in detail. But we **DO** choose our battles wisely!).

Then this SO post deals with calling fortran functions from python as well.

The next logical step was to look up the documentation for the python module ctypes. This talks about how the shared library can be accessed using python at an API level.

I had all the pieces to make a minimal working example, which another answer has already done. But i wanted to look at the output mechanism and mathematical operations involving real floats. Here is the test case i made.

`function prnt(s)`

character(80):: s

logical :: prnt

print*, s

prnt = .true.

end function prnt

function sin_2(r)

real:: r,sin_2

sin_2 = sin(r)**2

end function sin_2

`$gfortran -shared -g -o test.so test.f90`

EDIT: for some reason my work computer needs the -fPIC option to compile

To make sure my two functions

`prnt`

`sin_2`

`nm`

`$ nm test.so | tail -3`

0000067f T prnt_

0000065c T sin_2_

U sinf@@GLIBC_2.0

So far so good. My functions

`prnt`

`sin_2`

`prnt_`

`sin_2_`

this is where all of this gets a bit soggy. Using the table in python-ctypes documentation, I typed in the following -

`>>> from ctypes import byref, cdll, c_float,c_char_p`

>>> t = cdll.LoadLibrary('./test.so')

>>> c = c_char_p("Mary had a little lamb")

>>> t.prnt_('Mary had a little lamb')

Mary had a little lambe

1

>>> t.prnt_("Mary had a little lamb")

Mary had a little lambe

1

>>> t.prnt_(c)

Mary had a little lambe[� .prnt_(c)

1

I suppose that the 1 printed at the end of each output is python's way of letting me know that the boolean output to

`t.prnt_`

`.true.`

I am a bit worried about how the situation gets worse with

`t.prnt_`

`e`

Then there is the

`t.sin_2_`

`>>> f = c_float(4.56)`

>>> t.sin_2_(4.56)

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

ctypes.ArgumentError: argument 1: <type 'exceptions.TypeError'>: Don't know how to convert parameter 1

>>> t.sin_2_(f)

Segmentation fault (core dumped)

Where did i go wrong here? I tried to explain how I approached the problem so that it would be apparent if I made an obvious gaffe somewhere.

The multitude of links to other SO posts is to help other people who ask the same question as I do now.

Answer

In Fortran, arguments are passed by reference. Fortran character arrays aren't null terminated; the length is passed by value as an implicit `long int`

argument. Also, Python's `float`

type is a `double`

, so you may want to use Fortran `real(8)`

for consistency.

test.f90:

```
function prnt(s) ! byref(s), byval(length) [long int, implicit]
character(len=*):: s ! variable length input
logical :: prnt
write(*, "(A)") s ! formatted, to remove initial space
prnt = .true.
end function prnt
function sin_2(r) ! byref(r)
real:: r, sin_2 ! float; use real(8) for double
sin_2 = sin(r)**2
end function sin_2
```

Remember to set ctypes `argtypes`

for functions, and `restype`

where appropriate. In this case `sin_2`

takes a float pointer and returns a float.

ctypes example:

```
>>> from ctypes import *
>>> test = CDLL('./test.so')
>>> test.prnt_.argtypes = [c_char_p, c_long]
>>> test.sin_2_.argtypes = [POINTER(c_float)]
>>> test.sin_2_.restype = c_float
>>> s = 'Mary had a little lamb'
>>> test.prnt_(s, len(s))
Mary had a little lamb
1
>>> x = c_float(4.56)
>>> test.sin_2_(byref(x))
0.9769567847251892
```

Source (Stackoverflow)

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