What happens if you write a variable name alone in python?

Question

Recently I became curious about but what happens in line 2 of the following bogus python code:

def my_fun(foo,bar):
    foo
    return foo + bar

The reason I became interested is that I'm trying Light Table and tried to put a watch on "foo." It appeared to cause the python interpreter to hang.

Am I correct in thinking that this line has absolutely no effect and does not cause any sort of error? Can someone explain what the interpreter does exactly here?

Answer 1

One can look at what is happening with a little help from the built-in dis module:

import dis

def my_fun(foo,bar):
    foo
    return foo + bar

dis.dis(my_fun)

The dis.dis function disassembles functions (yep, it can disassemble itself), methods, and classes.

The output of dis.dis(my_fun) is:

  4           0 LOAD_FAST                0 (foo)
              3 POP_TOP

  5           4 LOAD_FAST                0 (foo)
              7 LOAD_FAST                1 (bar)
             10 BINARY_ADD
             11 RETURN_VALUE

The first two bytecodes are exactly what we need: the foo line.

Here's what these bytecodes do:

1. The first one pushes a reference to a local variable foo onto the stack (LOAD_FAST)
2. The second one removes the top of the stack (POP_TOP)

Basically, foo line has no effect. (well, if foo variable is not defined then LOAD_FAST will throw the NameError)