I am participating in a CodeForces competition tomorrow, and the rules say that Python is compiled with the following line (where %1 is the filename):
python -c "compile(open('%1').read(), '%1', 'exec')"
I tried to compile a test file with this line, but it does not do anything at all:
import sys
a = sys.stdin.readline()
sys.stdout.write(a)
However, the program works when I compile with python test.py
How can I make this test file work with the compilation line above?
EDIT: I am using terminal on a mac.
You can see what is happening if you try it in the interactive interpreter:
>>> compile(open('test.py').read(), 'read.py', 'exec')
<code object <module> at 0x10b916130, file "read.py", line 1>
The compile built-in compiles the source lines into a code object. To actually run the code object, you need to exec it:
>>> codeobj = compile(open('test.py').read(), 'read.py', 'exec')
>>> exec(codeobj)
Hello, world!
Hello, world!
>>>
Note that there are some differences here between Python 2 and Python 3, primarily that exec is a statement in Py2 but a built-in function in Py3. The above should work in either.