Python 函数参数问题

Python的传递究竟是值传递还是引用传递？

在回答这个问题之前，需要知道python中的变量只是一个对象的引用。赋值操作不会改变对象指向的内容，而是把变量从一个对象的引用，改为指向另一个对象的引用。对一个变量重新复制后，复制前后的变量id会发生改变。而在变量直接进行操作情况，分可变对象（list dict...）和不可变对象（int str tuple...）。前者不会改变，而后者则会改变，由于原始的不能改变，只能返回一个新值并赋值给该变量。

In [1]: i = 1

In [2]: id(i)
Out[2]: 31122120L

In [3]: i = 5

In [4]: id(i)
Out[4]: 31122024L

In [5]: a = [1,2,3]

In [6]: id(a)
Out[6]: 62749960L

In [7]: a = [23]

In [8]: id(a)
Out[8]: 62757576L

In [9]: i = 5

In [10]: id(i)
Out[10]: 31122024L

In [11]: i += 1

In [12]: id(i)
Out[12]: 31122000L

In [13]: id(a)
Out[13]: 62757576L

In [14]: a.append(5)

In [15]: id(a)
Out[15]: 62757576L

赋值无法改变函数内或者函数外，变量引用地址的对象。因此python函数是按引用传递，传递的是对象的引用。至于会不会修改这个参数，取决于对变量进行的操作和变量是 可变变量还是不可变变量。

def f(n,x):
    n = 3
    x.append(7)
    #x = [4,5,6]
    print "In f()",n,x

if __name__ == "__main__":

    n = 1
    x = [0,1,2]

    print "Before:",n,x
    f(n,x)
    print "After:",n,x

>>> 
Before: 1 [0, 1, 2]
In f() 3 [0, 1, 2, 7]
After: 1 [0, 1, 2, 7]

def f(n,x):
    n += 3
    x.append(7)
    #x = [4,5,6]
    print "In f()",n,x

if __name__ == "__main__":

    n = 1
    x = [0,1,2]

    print "Before:",n,x
    f(n,x)
    print "After:",n,x

Before: 1 [0, 1, 2]
In f() 4 [0, 1, 2, 7]
After: 1 [0, 1, 2, 7]

def f(n,x):
    n = 3
    #x.append(7)
    x = [4,5,6]
    print "In f()",n,x

if __name__ == "__main__":

    n = 1
    x = [0,1,2]

    print "Before:",n,x
    f(n,x)
    print "After:",n,x

Before: 1 [0, 1, 2]
In f() 3 [4, 5, 6]
After: 1 [0, 1, 2]