当Python 2.7.9中的Change Lists Global和Variable Global时，为什么会有所不同？ [重复]

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Referencing global primitives vs. objects 1 answer

引用全局原语与对象1回答

I have 2 example below, and my question is: why List L don't need global to change value like variable V, I think, more Safe if L need global to change

我下面有两个例子,我的问题是:为什么List L不需要全局来改变变量V这样的值,我认为,如果L需要全局改变则更安全

Example 1:

L=[1,2,3,4]  
V=5          
def U():
    L[1]=22
    V=55
U()
print L[1]
print V

Result 1:

22
5

Example 2:

L=[1,2,3,4]  
V=5          
def U():
    L[1]=22  
    global V   #Now V will change like L[1]
    V=55
U()
print L[1]
print V

Result 2:

22
55

1 个解决方案

#1

Perhaps someone can explain it better than me, but when you do the following:

也许有人可以比我更好地解释它,但当你做以下事情时:

print V

Python knows this variable isn't defined in your functions scope, and so it checks the global scope, and prints it.

Python知道这个变量没有在你的函数范围中定义,因此它会检查全局范围并打印它。

If you were to do instead

如果你要这样做

V=5

Python knows this variable isn't defined in your functions scope, so it goes ahead and creates it (now it over-shadows V at the global scope). There is different behaviour between writing a variable and reading it.

Python知道这个变量没有在你的函数范围中定义,因此它继续并创建它(现在它在全局范围内覆盖了V)。在编写变量和读取变量之间存在不同的行为。

Now compare this to a list:

现在将其与列表进行比较:

L[1]=22

This is similar to print V. Why? Because Python needs to find where L is defined in memory, and then modifies its second index. This is a read followed by a write, whereas print V is simply a read from memory.

这与print V类似。为什么?因为Python需要找到L在内存中定义的位置,然后修改其第二个索引。这是一个读取后写入,而打印V只是从内存中读取。

When your operation needs to do a read first, it will find the reference that was defined at the outer scope, and that is why you see a difference between V=55 and L[1]=22.

当您的操作需要先读取时,它将找到在外部范围定义的引用,这就是您看到V = 55和L [1] = 22之间的差异的原因。

When your operation is a write first, it will create a new variable in the function scope unless you have that variable marked as a global. Then it will modify that global instead.

当您的操作首先写入时,它将在函数范围中创建一个新变量,除非您将该变量标记为全局变量。然后它将修改该全局。

#1