如何更改函数中变量的作用域?Python [英] How to change the scope of a variable in a function? Python

问题描述

这似乎是一个非常愚蠢的问题，但我对 Python 中的范围规则感到困惑.在下面的示例中，我将两个带有值的变量 (x,y) 发送到一个应该更改其值的函数.当我打印结果时，变量没有改变.

This may seem like a really stupid question but I am confused regarding the scope rules in Python. In the following example I send two variables (x,y) with values to a function which is supposed to change their values. When I print the results the variables had not changed.

def func1(x,y):
    x=200
    y=300

x=2
y=3

func1(x,y)

print x,y #prints 2,3

现在，如果这是 C++，我会通过引用 (&) 将它们发送到该函数，因此能够更改它们的值.那么 Python 中的等效项是什么?更重要的是，当您将对象发送到函数时实际发生了什么?Python 是否会对这些对象进行新的引用?

Now if this were C++ I would send them by reference (&) to that function and therefore be able to change their values. So what's the equivilant in Python? and more important, what actually happens when you send objects to function? does Python make new references to these objects?

推荐答案

将它们视为功能的一部分.当函数结束时，它的所有变量也会消失.

Think of them as being part of the function. When the function ends, all its variables die too.

x=2
y=3

def func(x,y):
    x=200
    y=300

func(x,y) #inside this function, x=200 and y=300
#but by this line the function is over and those new values are discarded
print(x,y) #so this is looking at the outer scope again

如果您希望函数完全按照您编写的方式修改值，您可以使用 global，但这是非常糟糕的做法.

If you want a function to modify a value in exactly the way you have written it, you could use a global but this is VERY bad practice.

def func(x,y):
    global x #these tell the function to look at the outer scope 
    global y #and use those references to x and y, not the inner scope
    x=200
    y=300

func(x,y)
print(x,y) #prints 200 300

这样做的问题在于，它在最好的情况下使调试成为一场噩梦，而在最坏的情况下则完全无法理解.像这样的事情在函数中通常被称为副作用"——设置一个不需要设置的值并且这样做而不明确返回它是一种坏事.通常，您应该编写的唯一可以就地修改项目的函数是对象方法(例如 [].append() 修改列表，因为返回一个新列表是愚蠢的！)

The problem with this is that it makes debugging a nightmare in the best case, and utterly incomprehensibly impossible in the worst case. Things like these are commonly known as "side effects" in functions -- setting a value you don't need set and doing so without explicitly returning it is kind of a Bad Thing. Generally the only functions you should write that modify items in-place are object methods (things like [].append() modify the list because it's silly to return a new list instead!)

执行此类操作的正确方法是使用返回值.尝试类似

The RIGHT way to do something like this would be to use a return value. Try something like

def func(x,y):
    x = x+200 #this can be written x += 200
    y = y+300 #as above: y += 300
    return (x,y) #returns a tuple (x,y)

x = 2
y = 3
func(x,y) # returns (202, 303)
print(x,y) #prints 2 3

为什么那不起作用?嗯，因为你从来没有告诉程序对那个元组 (202, 303) 做任何事情，只是为了计算它.让我们现在分配它

Why didn't that work? Well because you never told the program to DO anything with that tuple (202, 303), just to calculate it. Let's assign it now

#func as defined above

x=2 ; y=3
x,y = func(x,y) #this unpacks the tuple (202,303) into two values and x and y
print(x,y) #prints 202 303

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