全局和局部变量在这种情况下的表现如何？ [英] How do global and local variables behave in this case?

问题描述

保存在我的文件 function_local.py ：

x = 50

def func(x):
    print('x is', x)
    x = 2
    print('Changed local x to', x)

func(x)
print('x is still', x)

输出：

Output:

$ python function_local.py
x is 50
Changed local x to 2
x is still 50

问题：当我们在Function中打印第一行时，为什么它打印出50，而不是2？即使这样说，在函数中， x = 2 ？

Question: When we print the first line inside the Function, why does it print out 50, not 2? Even if it is said, in the function, that x = 2?

推荐答案

如果您在函数中分配了一个变量名称（未声明 global 或 nonlocal ）或者在函数的参数列表中使用变量名称，变量名称将成为函数的一部分。

In case you assign to a variable name (that wasn't declared global or nonlocal) in a function or use the variable name in the argument list of the function the variable name will become part of the function.

在这种情况下，您可以在函数内使用任何变量名称，因为它始终会引用传入的局部变量：

In that case you could've used any variable name inside the function because it will always refer to the local variable that was passed in:

x = 50

def func(another_x):
    print('local x =', another_x)
    another_x = 2
    print('local x =', another_x)
    return another_x

print('global x =', x)
x = func(x)                # note that I assigned the returned value to "x" to make the change visible outside
print('global x =', x)

更多解释

我会试着展示我以前的意思，当我说 x with将成为函数的一部分。

More explanation

I'm going to try to show what I meant earlier when I said that x with "will become part of the function".

__ code __。co_varnames 包含函数的局部变量名称列表。所以让我们看看在几种情况下会发生什么：

The __code__.co_varnames of a function holds the list of local variable names of the function. So let's see what happens in a few cases:

如果它是签名的一部分：

If it's part of the signature:

def func(x):    # the name "x" is part of the argument list
    pass

print(func.__code__.co_varnames)
# ('x',)

如果分配给变量（函数中的任何位置） p>

If you assign to the variable (anywhere in the function):

def func():
    x = 2    # assignment to name "x" inside the function

print(func.__code__.co_varnames)
# ('x',)

您只需访问变量名称：

If you only access the variable name:

def func():
    print(x)

print(func.__code__.co_varnames)
# ()

在这种情况下，它实际上将查找外部作用域中的变量名称，因为变量名 x 不是函数varnames的一部分！

In this case it will actually look up the variable name in the outer scopes because the variable name x isn't part of the functions varnames!

我可以理解这会如何混淆你，因为只需添加一个 x =

I can understand how this could confuse you because just by adding a x=<whatever> anywhere in the function will make it part of the function:

def func():
    print(x)  # access
    x = 2     # assignment

print(func.__code__.co_varnames)
# ('x',)

在这种情况下，不会查找变量 x ，因为它是现在函数的一部分，您将得到一个告诉异常（因为即使 x 是

In that case it won't look up the variable x from the outer scope because it's part of the function now and you'll get a tell-tale Exception (because even though x is part of the function it isn't assigned to yet):

>>> func()
UnboundLocalError: local variable 'x' referenced before assignment

显然，它会如果您在分配给它后访问它：

Obviously, it would work if you access it after assigning to it:

def func():
    x = 2     # assignment
    print(x)  # access

或者如果您在调用函数时将其传入：

Or if you pass it in when you call the function:

def func(x):  # x will be passed in
    print(x)  # access

关于局部变量名称的另一个重点是您不能从外部范围设置变量，除非您明确告诉Python使用全局或非本地


Another important point about local variable names is that you can't set variables from outer scopes, except when you tell Python to explicitly not make x part of the local variable names, for example with global or nonlocal:

def func():
    global x  # or "nonlocal x"
    print(x)
    x = 2

print(func.__code__.co_varnames)
# ()

这实际上会覆盖全球（o当你调用 func（）！

This will actually overwrite what the global (or nonlocal) variable name x refers to when you call func()!

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