python - "UnboundLocalError"报告错误的行号

Question

例如:

$ cat -n foo.py
     1  def f():
     2      str = len
     3      str = str('abc')
     4  #   len = len('abc')
     5  f()
$ python2.7 foo.py
$

它运行成功，因此第 2 行和第 3 行没有问题。但在我取消注释第 4 行之后:

$ cat -n bar.py
     1  def f():
     2      str = len
     3      str = str('abc')
     4      len = len('abc')
     5  f()
$ python2.7 bar.py
Traceback (most recent call last):
  File "bar.py", line 5, in <module>
    f()
  File "bar.py", line 2, in f
    str = len
UnboundLocalError: local variable 'len' referenced before assignment
$

现在它报告错误，因此未注释的第 4 行肯定有问题，但为什么第 2 行报告回溯错误？

Answer 1

编程常见问题解答中有答案

This is because when you make an assignment to a variable in a scope, that variable becomes local to that scope and shadows any similarly named variable in the outer scope.

在此处阅读完整内容:Why am I getting an UnboundLocalError when the variable has a value?

当 len 被注释时，它被视为内置函数 len()

def f():
    str = len
    print type(str)
    str = str('abc')
    # len = len('abc')
    print type(len)

f()

<type 'builtin_function_or_method'>
<type 'builtin_function_or_method'>