通过引用调用 Python 函数

In Python the passing by reference or by value has to do with what are the actual objects you are passing.So,if you are passing a list for example,then you actually make this pass by reference,since the list is a mutable object.Thus,you are passing a pointer to the function and you can modify the object (list) in the function body.

When you are passing a string,this passing is done by value,so a new string object is being created and when the function terminates it is destroyed. So it all has to do with mutable and immutable objects.

You can not change an immutable object, like str or tuple, inside a function in Python, but you can do things like:

def foo(y):
y[0] = y[0]**2


x = [5]
foo(x)
print x[0]  # prints 25

That is a weird way to go about it, however, unless you need to always square certain elements in an array.

Note that in Python, you can also return more than one value, making some of the use cases for pass by reference less important:

def foo(x, y):
return x**2, y**2


a = 2
b = 3
a, b = foo(a, b)  # a == 4; b == 9

When you return values like that, they are being returned as a Tuple which is in turn unpacked.

edit: Another way to think about this is that, while you can't explicitly pass variables by reference in Python, you can modify the properties of objects that were passed in. In my example (and others) you can modify members of the list that was passed in. You would not, however, be able to reassign the passed in variable entirely. For instance, see the following two pieces of code look like they might do something similar, but end up with different results:

def clear_a(x):
x = []


def clear_b(x):
while x: x.pop()


z = [1,2,3]
clear_a(z) # z will not be changed
clear_b(z) # z will be emptied

So this is a little bit of a subtle point, because while Python only passes variables by value, every variable in Python is a reference. If you want to be able to change your values with a function call, what you need is a mutable object. For example:

l = [0]


def set_3(x):
x[0] = 3


set_3(l)
print(l[0])

In the above code, the function modifies the contents of a List object (which is mutable), and so the output is 3 instead of 0.

I write this answer only to illustrate what 'by value' means in Python. The above code is bad style, and if you really want to mutate your values you should write a class and call methods within that class, as MPX suggests.

The answer given is

def set_4(x):
y = []
for i in x:
y.append(i)
y[0] = 4
return y

and

l = [0]


def set_3(x):
x[0] = 3


set_3(l)
print(l[0])

which is the best answer so far as it does what it says in the question. However,it does seem a very clumsy way compared to VB or Pascal.Is it the best method we have?

Not only is it clumsy, it involves mutating the original parameter in some way manually eg by changing the original parameter to a list: or copying it to another list rather than just saying: "use this parameter as a value " or "use this one as a reference". Could the simple answer be there is no reserved word for this but these are great work arounds?

OK, I'll take a stab at this. Python passes by object reference, which is different from what you'd normally think of as "by reference" or "by value". Take this example:

def foo(x):
print x


bar = 'some value'
foo(bar)

So you're creating a string object with value 'some value' and "binding" it to a variable named bar. In C, that would be similar to bar being a pointer to 'some value'.

When you call foo(bar), you're not passing in bar itself. You're passing in bar's value: a pointer to 'some value'. At that point, there are two "pointers" to the same string object.

Now compare that to:

def foo(x):
x = 'another value'
print x


bar = 'some value'
foo(bar)

Here's where the difference lies. In the line:

x = 'another value'

you're not actually altering the contents of x. In fact, that's not even possible. Instead, you're creating a new string object with value 'another value'. That assignment operator? It isn't saying "overwrite the thing x is pointing at with the new value". It's saying "update x to point at the new object instead". After that line, there are two string objects: 'some value' (with bar pointing at it) and 'another value' (with x pointing at it).

This isn't clumsy. When you understand how it works, it's a beautifully elegant, efficient system.

Hope the following description sums it up well:

There are two things to consider here - variables and objects.

1. If you are passing a variable, then it's pass by value, which means the changes made to the variable within the function are local to that function and hence won't be reflected globally. This is more of a 'C' like behavior.

Example:

def changeval( myvar ):
myvar = 20;
print "values inside the function: ", myvar
return


myvar = 10;
changeval( myvar );
print "values outside the function: ", myvar

O/P:

values inside the function:  20
values outside the function:  10

2. If you are passing the variables packed inside a mutable object, like a list, then the changes made to the object are reflected globally as long as the object is not re-assigned.

Example:

def changelist( mylist ):
mylist2=['a'];
mylist.append(mylist2);
print "values inside the function: ", mylist
return


mylist = [1,2,3];
changelist( mylist );
print "values outside the function: ", mylist

O/P:

values inside the function:  [1, 2, 3, ['a']]
values outside the function:  [1, 2, 3, ['a']]

3. Now consider the case where the object is re-assigned. In this case, the object refers to a new memory location which is local to the function in which this happens and hence not reflected globally.

Example:

def changelist( mylist ):
mylist=['a'];
print "values inside the function: ", mylist
return


mylist = [1,2,3];
changelist( mylist );
print "values outside the function: ", mylist

O/P:

values inside the function:  ['a']
values outside the function:  [1, 2, 3]

Python is neither pass-by-value nor pass-by-reference. It's more of "object references are passed by value" as described here:

1. Here's why it's not pass-by-value. Because

   def append(list):
   list.append(1)
   
   
   list = [0]
   reassign(list)
   append(list)
   

returns [0,1] showing that some kind of reference was clearly passed as pass-by-value does not allow a function to alter the parent scope at all.

Looks like pass-by-reference then, hu? Nope.

2. Here's why it's not pass-by-reference. Because

   def reassign(list):
   list = [0, 1]
   
   
   list = [0]
   reassign(list)
   print list
   

returns [0] showing that the original reference was destroyed when list was reassigned. pass-by-reference would have returned [0,1].

For more information look here:

If you want your function to not manipulate outside scope, you need to make a copy of the input parameters that creates a new object.

from copy import copy


def append(list):
list2 = copy(list)
list2.append(1)
print list2


list = [0]
append(list)
print list

There are essentially three kinds of 'function calls':

• Pass by value
• Pass by reference
• Pass by object reference

Python is a PASS-BY-OBJECT-REFERENCE programming language.

Firstly, it is important to understand that a variable, and the value of the variable (the object) are two seperate things. The variable 'points to' the object. The variable is not the object. Again:

THE VARIABLE IS NOT THE OBJECT

Example: in the following line of code:

>>> x = []

[] is the empty list, x is a variable that points to the empty list, but x itself is not the empty list.

Consider the variable (x, in the above case) as a box, and 'the value' of the variable ([]) as the object inside the box.

PASS BY OBJECT REFERENCE (Case in python):

Here, "Object references are passed by value."

def append_one(li):
li.append(1)
x = [0]
append_one(x)
print x

Here, the statement x = [0] makes a variable x (box) that points towards the object [0].

On the function being called, a new box li is created. The contents of li are the SAME as the contents of the box x. Both the boxes contain the same object. That is, both the variables point to the same object in memory. Hence, any change to the object pointed at by li will also be reflected by the object pointed at by x.

In conclusion, the output of the above program will be:

[0, 1]

Note:

If the variable li is reassigned in the function, then li will point to a separate object in memory. x however, will continue pointing to the same object in memory it was pointing to earlier.

Example:

def append_one(li):
li = [0, 1]
x = [0]
append_one(x)
print x

The output of the program will be:

[0]

PASS BY REFERENCE:

The box from the calling function is passed on to the called function. Implicitly, the contents of the box (the value of the variable) is passed on to the called function. Hence, any change to the contents of the box in the called function will be reflected in the calling function.

PASS BY VALUE:

A new box is created in the called function, and copies of contents of the box from the calling function is stored into the new boxes.

Hope this helps.

Python already call by ref..

let's take example:

  def foo(var):
print(hex(id(var)))




x = 1 # any value
print(hex(id(x))) # I think the id() give the ref...
foo(x)

OutPut

  0x50d43700 #with you might give another hex number deppend on your memory
0x50d43700

Consider that the variable is a box and the value it points to is the "thing" inside the box:

1. Pass by reference : function shares the same box and thereby the thing inside also.

2. Pass by value : function creates a new box, a replica of the old one, including a copy of whatever thing is inside it. Eg. Java - functions create a copy of the box and the thing inside it which can be: a primitive / a reference to an object. (note that the copied reference in the new box and the original both still point to the same object, here the reference IS the thing inside the box, not the object it is pointing to)

3. Pass by object-reference: the function creates a box, but it encloses the same thing the initial box was enclosing. So in Python:

a) if the thing inside said box is mutable, changes made will reflect back in the original box (eg. lists)

b) if the thing is immutable (like python strings and numeric types), then the box inside the function will hold the same thing UNTIL you try to change its value. Once changed, the thing in the function's box is a totally new thing compared to the original one. Hence id() for that box will now give the identity of the new thing it encloses.

Technically python do not pass arguments by value: all by reference. But ... since python has two types of objects: immutable and mutable, here is what happens:

• Immutable arguments are effectively passed by value: string, integer, tuple are all immutable object types. While they are technically "passed by reference" (like all parameters), since you can't change them in-place inside the function it looks/behaves as if it is passed by value.

• Mutable arguments are effectively passed by reference: lists or dictionaries are passed by its pointers. Any in-place change inside the function like (append or del) will affect the original object.

This is how Python is designed: no copies and all are passed by reference. You can explicitly pass a copy.

def sort(array):
# do sort
return array


data = [1, 2, 3]
sort(data[:]) # here you passed a copy

Last point I would like to mention which is a function has its own scope.

def do_any_stuff_to_these_objects(a, b):
a = a * 2
del b['last_name']


number = 1 # immutable
hashmap = {'first_name' : 'john', 'last_name': 'legend'} # mutable
do_any_stuff_to_these_objects(number, hashmap)
print(number) # 1 , oh  it should be 2 ! no a is changed inisde the function scope
print(hashmap) # {'first_name': 'john'}

class demoClass:
x = 4
y = 3
foo1 = demoClass()
foo1.x = 2
foo2 = demoClass()
foo2.y = 5
def mySquare(myObj):
myObj.x = myObj.x**2
myObj.y = myObj.y**2
print('foo1.x =', foo1.x)
print('foo1.y =', foo1.y)
print('foo2.x =', foo2.x)
print('foo2.y =', foo2.y)
mySquare(foo1)
mySquare(foo2)
print('After square:')
print('foo1.x =', foo1.x)
print('foo1.y =', foo1.y)
print('foo2.x =', foo2.x)
print('foo2.y =', foo2.y)