在 Python 中通过引用传递整数 [英] Passing an integer by reference in Python
问题描述
如何在 Python 中通过引用传递整数?
How can I pass an integer by reference in Python?
我想修改传递给函数的变量的值.我读过 Python 中的所有内容都是按值传递的，但必须有一个简单的技巧.例如，在 Java 中，您可以传递 Integer、Long 等的引用类型.
I want to modify the value of a variable that I am passing to the function. I have read that everything in Python is pass by value, but there has to be an easy trick. For example, in Java you could pass the reference types of Integer, Long, etc.
- 如何通过引用将整数传递给函数?
- 最佳做法是什么?
推荐答案
它在 Python 中不太适用.Python 传递对对象的引用.在您的函数中，您有一个对象——您可以随意修改该对象(如果可能).然而，整数是不可变的.一种解决方法是将整数传递到可以变异的容器中:
It doesn't quite work that way in Python. Python passes references to objects. Inside your function you have an object -- You're free to mutate that object (if possible). However, integers are immutable. One workaround is to pass the integer in a container which can be mutated:
def change(x):
x[0] = 3
x = [1]
change(x)
print x
这充其量是丑陋/笨拙的，但你不会在 Python 中做得更好.原因是因为在 Python 中，赋值 (=) 将任何对象作为右侧的结果并将其绑定到左侧的任何对象 *(或将其传递给适当的函数).
This is ugly/clumsy at best, but you're not going to do any better in Python. The reason is because in Python, assignment (=) takes whatever object is the result of the right hand side and binds it to whatever is on the left hand side *(or passes it to the appropriate function).
理解了这一点，我们就可以明白为什么没有办法改变函数内不可变对象的值——你不能改变它的任何属性，因为它是不可变的，你不能只分配变量"" 一个新值，因为这样您实际上是在创建一个新对象(与旧对象不同)并为其赋予旧对象在本地命名空间中的名称.
Understanding this, we can see why there is no way to change the value of an immutable object inside a function -- you can't change any of its attributes because it's immutable, and you can't just assign the "variable" a new value because then you're actually creating a new object (which is distinct from the old one) and giving it the name that the old object had in the local namespace.
通常的解决方法是简单地返回您想要的对象:
Usually the workaround is to simply return the object that you want:
def multiply_by_2(x):
return 2*x
x = 1
x = multiply_by_2(x)
<小时>
*在上面的第一个例子中，3 实际上被传递给了 x.__setitem__.
*In the first example case above, 3 actually gets passed to x.__setitem__.
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