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面试题之一。
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：
我以为像这种传参作为局部变量，因为都不会影响外部的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] >>>
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，可以看到，全局变量中的id(x) = 46798728，x进到func()中，因为执行了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])对象时如何变化的吧：
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()估计也是。
利用id(x)，查看下x.extend([1])对象时如何变化的吧：
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，他们的区别大概就是：
>>> 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：
源代码地址：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 p2p2传入func2()，因为+=操作，就是list.extend()，操作，在原对象操作，所以p2=[1,2,3,1]。
吐槽下：
其实python在函数中参数是传引用的，如果一个对象obj进到函数中，被改变，那无论在哪里这个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，因为这个被改变后的obj不是原来的它了。比如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）。
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python list的+，+=，append，extend
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原文地址：http://blog.csdn.net/emaste_r/article/details/47395055
