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 p1print 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 = ",xfunc(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) = 46798728in func(), local! original x = [1, 2, 3] id(x) = 46798728in func(), local! now x = [1] id(x) = 46789512after 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) = 46339976in func(), local! original x = [1, 2, 3] id(x) = 46339976in func(), local! now x = [1, 2, 3, 1] id(x) = 46390664after 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) = 46536584in func(), local! original x = [1, 2, 3] id(x) = 46536584in func(), local! now x = [1, 2, 3, 1] id(x) = 46536584after 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) = 47191944in func(), local! original x = [1, 2, 3] id(x) = 47191944in func(), local! now x = [1, 2, 3, [1]] id(x) = 47191944after 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) = 48437128in func(), local! original x = [1, 2, 3] id(x) = 48437128in func(), local! now x = [1, 2, 3, 1] id(x) = 48437128after 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
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) = 47781768in func(), local! original x = [1, 2, 3] id(x) = 47781768in func(), local! now x = [1, 2, 3, 1] id(x) = 47795720after 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 p1print p2p1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）。