scope of eval function in python

Question

Consider the following example:

i=7
j=8
k=10
def test():
    i=1
    j=2
    k=3
    return dict((name,eval(name)) for name in [\'i\',\'j\',\'k\'])
         


        

Answer 1

This occurs because the generator expression has a different scope to the function:

>>> def test():
    i, j, k = range(1, 4)
    return dict((j, locals()) for _ in range(i))

>>> test()
{2: {'.0': , 'j': 2, '_': 0}}

Using j inside the scope binds it from the function, as that's the nearest enclosing scope, but i and k are not locally bound (as k isn't referenced and i is only used to create the range).

---

Note that you can avoid this issue with:

return dict(i=i, j=j, k=k)

or a dictionary literal:

return {'i': i, 'j': j, 'k': k}