封闭的局部范围内的变量与全局范围内的变量-Python

Question

hi guys I can't figure out why when find_average() is called, total = 20 in the global scope is being used for the find_total() function, instead of total = 10 in the enclosing scope? Thanks in advance for any insights and help!

total = 20

def find_total(l):
    return total

def find_length(l):
    length = len(l)
    return length

def find_average(l):
    total = 10
    return find_total(l) / find_length(l)


average = find_average(example_list)

Answer 1

Each function has its own scope. It starts at the functions local, inner scope, then goes outwards through enclosing functions until it reaches global (module) scope. This sequence depends on the scope the function is defined in . The stack (calling sequence) is not used for variable lookup.

In your example, each function only has its inner scope followed by the outer scope. For find_total , that's <module>.find_total.<locals> and <module> . So, whenever find_total is run, it will look up total in its local scope, failing, and look in the global scope. There, total == 20 .

The scope inside find_average is exclusive to find_average . Neither global scope, find_total or find_length can access it. If you want to pass something from inside find_average to find_total , you should do so via a parameter.

---

Alternatively, if find_total can be defined inside find_average . This way, find_total resolves names by searching the sequence <module>.find_average.<locals>.find_total.<locals> -> <module>.find_average.<locals> -> <module> .

total = 20

def find_length(l):
  length = len(l)
  return length

def find_average(l):
  total = 10
  def find_total(l):
    return total
  return find_total(l) / find_length(l)

 average = find_average(example_list)

This will make find_total inaccessible from outside find_average !