Inspect lambda code from Python interpreter

Question

this simple python program that is extracted from more complex codebase:

#insp.py
import inspect
L = lambda x: x+1
print("L(10)=" + str(L(10)))
code = inspect.getsource(L)
print(code)

works if i run it from the command line as:

$ python insp.py

If I copy and paste each line in the python interpreter it fails:

d:\>python
Python 3.5.1 (v3.5.1:37a07cee5969, Dec  6 2015, 01:38:48) [MSC v.1900 32 bit (Intel)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import inspect
>>> L = lambda x: x+1
>>> print("L(10)=" + str(L(10)))
L(10)=11
>>> code = inspect.getsource(L)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "d:\Users\Cimino\AppData\Local\Programs\Python\Python35-32\Lib\inspect.py", line 944, in getsource
    lines, lnum = getsourcelines(object)
  File "d:\Users\Cimino\AppData\Local\Programs\Python\Python35-32\Lib\inspect.py", line 931, in getsourcelines
    lines, lnum = findsource(object)
  File "d:\Users\Cimino\AppData\Local\Programs\Python\Python35-32\Lib\inspect.py", line 762, in findsource
    raise OSError('could not get source code')
OSError: could not get source code

Note that using IPython instead of the plain Python interpreter, it works!

Does anybody know why?

I use Python 3.5 32 bits under Windows7.

Answer 1

It works in IPython because it caches each and every command you enter there using linecache module.

For example:

$ ipy  ## Equivalent to ipython --classic
Python 2.7.10 (default, Jul 30 2016, 18:31:42)
Type "copyright", "credits" or "license" for more information.

IPython 3.0.0 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help      -> Python's own help system.
object?   -> Details about 'object', use 'object??' for extra details.
>>> print a
Traceback (most recent call last):
  File "<ipython-input-1-9d7b17ad5387>", line 1, in <module>
    print a
NameError: name 'a' is not defined

Notice the <ipython-input-1-9d7b17ad5387> part here, this is something specific to IPython. In normal Python shell you would see <stdin>:

$ python
Python 2.7.10 (default, Jul 30 2016, 18:31:42)
[GCC 4.2.1 Compatible Apple LLVM 8.0.0 (clang-800.0.34)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> print a
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'a' is not defined

Now let's run your code:

>>> import inspect
>>> L = lambda x: x+1
>>> code = inspect.getsource(L)

Time to find out the filename related to L:

>>> L.func_code.co_filename
'<ipython-input-2-0c0d6f325784>'

Now let's see if we have the source in linecache.cache for this file:

>>> import linecache
>>> linecache.cache[L.func_code.co_filename]
(18, 1481047125.479239, [u'L = lambda x: x+1\n'], '<ipython-input-2-0c0d6f325784>')

So, using this information IPython is able to find the required source but Python shell doesn't because it is not storing any.

---

The related details about how inspect gets the source can be found in getsourcefile and findsource functions in the source code.