面试题之中的一个。

def func1(p):

    p = p + [1]

def func2(p):

    p += [1]

p1 = [1,2,3]

p2 = [1,2,3]

func1(p1)

func2(p2)

print p1

print p2

结果：

我以为像这样的传參作为局部变量。由于都不会影响外部的list。所以答案应该是p1 =[1,2,3] ，p2=[1,2,3]，然而

>>>

[1, 2, 3]

[1, 2, 3, 1]

>>> 

重新被面试官虐杀，不甘心的我查找了python 局部变量：

x = [1,2,3]

def func(x):

    print "local! original x = ",x

    x  = [1]

    print "local! now x = ",x

func(x)

print "global! x = ",x

结果：

local! original x =  [1, 2, 3]

local! now x =  [1]

global! x =  [1, 2, 3]

没错啊。我还记得要用全局变量得加global x 之类的语句呢。

为了保险起见，加一个id()，查查看对象是不是同一个先：

x = [1,2,3]

print "before func(), global! x = ",x,"id(x) = ",id(x)

def func(x):

    print "in func(), local! original x = ",x,"id(x) = ",id(x)

    x  = [1]

    print "in func(), local! now x = ",x,"id(x) = ",id(x)

func(x)

print "after func(), global! x = ",x,"id(x) = ",id(x)

结果：

before func(), global! x =  [1, 2, 3] id(x) =  46798728

in func(), local! original x =  [1, 2, 3] id(x) =  46798728

in func(), local! now x =  [1] id(x) =  46789512

after func(), global! x =  [1, 2, 3] id(x) =  46798728

恩，能够看到，全局变量中的id(x) = 46798728，x进到func()中，由于运行了x = [1]，才变成id(x) = 46789512。（合情合理）

这也说明python的确是传引用入函数。（然并卵）

利用id(x)，查看下x = x + [1]对象是怎么变化的吧：

x = [1,2,3]

print "before func(), global! x = ",x,"id(x) = ",id(x)

def func(x):

    print "in func(), local! original x = ",x,"id(x) = ",id(x)

    x  = x + [1]

    print "in func(), local! now x = ",x,"id(x) = ",id(x)

func(x)

print "after func(), global! x = ",x,"id(x) = ",id(x)

结果：

before func(), global! x =  [1, 2, 3] id(x) =  46339976

in func(), local! original x =  [1, 2, 3] id(x) =  46339976

in func(), local! now x =  [1, 2, 3, 1] id(x) =  46390664

after func(), global! x =  [1, 2, 3] id(x) =  46339976

啊。x = x + [1]，是新建了一个对象，id(x) =  46390664。

利用id(x)，查看下x += [1]对象是怎么变化的吧：

x = [1,2,3]

print "before func(), global! x = ",x,"id(x) = ",id(x)

def func(x):

    print "in func(), local! original x = ",x,"id(x) = ",id(x)

    x += [1]

    print "in func(), local! now x = ",x,"id(x) = ",id(x)

func(x)

print "after func(), global! x = ",x,"id(x) = ",id(x)

结果：

before func(), global! x =  [1, 2, 3] id(x) =  46536584

in func(), local! original x =  [1, 2, 3] id(x) =  46536584

in func(), local! now x =  [1, 2, 3, 1] id(x) =  46536584

after func(), global! x =  [1, 2, 3, 1] id(x) =  46536584

啊，id(x)全程一样。x += [1]，python直接就在原对象上操作，还真是够懒的说。

利用id(x)，查看下x.append([1])对象时怎样变化的吧：

x = [1,2,3]

print "before func(), global! x = ",x,"id(x) = ",id(x)

def func(x):

    print "in func(), local! original x = ",x,"id(x) = ",id(x)

    x.append([1])

    print "in func(), local! now x = ",x,"id(x) = ",id(x)

func(x)

print "after func(), global! x = ",x,"id(x) = ",id(x)

结果：

before func(), global! x =  [1, 2, 3] id(x) =  47191944

in func(), local! original x =  [1, 2, 3] id(x) =  47191944

in func(), local! now x =  [1, 2, 3, [1]] id(x) =  47191944

after func(), global! x =  [1, 2, 3, [1]] id(x) =  47191944

啊，id(x)全程一样，看来list的属性方法都是在原对象上操作的吧，我记得list.sort()也是，待会要验证的list.extend()预计也是。

利用id(x)，查看下x.extend([1])对象时怎样变化的吧：

x = [1,2,3]

print "before func(), global! x = ",x,"id(x) = ",id(x)

def func(x):

    print "in func(), local! original x = ",x,"id(x) = ",id(x)

    x.extend([1])

    print "in func(), local! now x = ",x,"id(x) = ",id(x)

func(x)

print "after func(), global! x = ",x,"id(x) = ",id(x)

结果：

before func(), global! x =  [1, 2, 3] id(x) =  48437128

in func(), local! original x =  [1, 2, 3] id(x) =  48437128

in func(), local! now x =  [1, 2, 3, 1] id(x) =  48437128

after func(), global! x =  [1, 2, 3, 1] id(x) =  48437128

果然id(x)全程一样。

话说list.append()是追加，extend()是拓展，他们的差别大概就是：

>>> a = [1,2,3]

>>> b = [4,5,6]

>>> c = [7,8,9]

>>> a.append(b)

>>> a

[1, 2, 3, [4, 5, 6]]

>>> c.extend(b)

>>> c

[7, 8, 9, 4, 5, 6]

>>>

看了上面的几段代码，聪明的你应该也能看出来：

list1 += list2  等价于 list1.extend(list2)，这是证据：

源码地址：

view=markup">http://svn.python.org/view/python/trunk/Objects/listobject.c?

view=markup

913	static PyObject *

914	list_inplace_concat(PyListObject *self, PyObject *other)

915	{

916	    PyObject *result;

917

918	    result = listextend(self, other); //+=果然用了listextend()

919	    if (result == NULL)

920	        return result;

921	    Py_DECREF(result);

922	    Py_INCREF(self);

923	    return (PyObject *)self;

924	}

利用id(x)。查看下global x下。对象的变化吧：

x = [1,2,3]

print "before func(), global! x = ",x,"id(x) = ",id(x)

def func():

    global x

    print "in func(), local! original x = ",x,"id(x) = ",id(x)

    x = x + [1]

    print "in func(), local! now x = ",x,"id(x) = ",id(x)

func()

print "after func(), global! x = ",x,"id(x) = ",id(x)

结果：

before func(), global! x =  [1, 2, 3] id(x) =  47781768

in func(), local! original x =  [1, 2, 3] id(x) =  47781768

in func(), local! now x =  [1, 2, 3, 1] id(x) =  47795720

after func(), global! x =  [1, 2, 3, 1] id(x) =  47795720

啊，global就保证了，即使我的变量x在函数中指向对象变了，外部的x也会指向新的对象。

回到面试题：

def func1(p):

    p = p + [1]

def func2(p):

    p += [1]

p1 = [1,2,3]

p2 = [1,2,3]

func1(p1)

func2(p2)

print p1

print p2

p1传入func1()。由于+操作，生成一个新的对象。但没有return给外部的p1。所以外部的p1=[1,2,3]。

p2传入func2()，由于+=操作，就是list.extend()。操作。在原对象操作。所以p2=[1,2,3,1]。

吐槽下：

事实上python在函数中參数是传引用的，假设一个对象obj进到函数中，被改变，那不管在哪里这个obj就是被改变了。并没有什么副本什么的。

那为什么有些时候看起来。函数中obj怎么被改变，外部的obj都岿然不动，啊，由于这个被改变后的obj不是原来的它了。

比方x = x + [1]。新的x真的是原来传进来的x吗？不是的。

此时的x是新的对象了（看id就知道了）。先前传进来的x。并没有被改变。

一点浅见，求打脸。

总结：

1、list + 创建一个新的对象。

2、list的 += 和 list.extend()，等价。都是在原对象上操作。

3、list.append()。也是在原对象上操作。

4、global，全局变量，嗯，不错（这算什么总结嘛）。