list res adding 試題 content int none tro append()

面試題之中的一個。

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

結果：

我以為像這樣的傳參作為局部變量。由於都不會影響外部的list。所以答案應該是p1 =[1,2,3] ，p2=[1,2,3]，然而

>>> 
[1, 2, 3]
[1, 2, 3, 1]
>>> 

重新被面試官虐殺，不甘心的我查找了python 局部變量：

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

結果：

local! original x =  [1, 2, 3]
local! now x =  [1]
global! x =  [1, 2, 3]

沒錯啊。我還記得要用全局變量得加global x 之類的語句呢。

為了保險起見，加一個id()，查查看對象是不是同一個先：

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)

結果：

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，能夠看到，全局變量中的id(x) = 46798728，x進到func()中，由於運行了x = [1]，才變成id(x) = 46789512。（合情合理）

這也說明python的確是傳引用入函數。（然並卵）

利用id(x)，查看下x = x + [1]對象是怎麽變化的吧：

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)

結果：

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]，是新建了一個對象，id(x) = 46390664。

利用id(x)，查看下x += [1]對象是怎麽變化的吧：

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)

結果：

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)全程一樣。x += [1]，python直接就在原對象上操作，還真是夠懶的說。

利用id(x)，查看下x.append([1])對象時怎樣變化的吧：

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)

結果：

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)全程一樣，看來list的屬性方法都是在原對象上操作的吧，我記得list.sort()也是，待會要驗證的list.extend()預計也是。

利用id(x)，查看下x.extend([1])對象時怎樣變化的吧：

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)

結果：

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)全程一樣。

話說list.append()是追加，extend()是拓展，他們的差別大概就是：

>>> 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]
>>>

看了上面的幾段代碼，聰明的你應該也能看出來：

list1 += list2 等價於 list1.extend(list2)，這是證據：

源碼地址：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。對象的變化吧：

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)

結果：

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就保證了，即使我的變量x在函數中指向對象變了，外部的x也會指向新的對象。

回到面試題：

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傳入func1()。由於+操作，生成一個新的對象。但沒有return給外部的p1。所以外部的p1=[1,2,3]。

p2傳入func2()，由於+=操作，就是list.extend()。操作。在原對象操作。所以p2=[1,2,3,1]。

吐槽下：

事實上python在函數中參數是傳引用的，假設一個對象obj進到函數中，被改變，那不管在哪裏這個obj就是被改變了。並沒有什麽副本什麽的。

那為什麽有些時候看起來。函數中obj怎麽被改變，外部的obj都巋然不動，啊，由於這個被改變後的obj不是原來的它了。

比方x = x + [1]。新的x真的是原來傳進來的x嗎？不是的。

此時的x是新的對象了（看id就知道了）。先前傳進來的x。並沒有被改變。

一點淺見，求打臉。

總結：

1、list + 創建一個新的對象。

2、list的 += 和 list.extend()，等價。都是在原對象上操作。

3、list.append()。也是在原對象上操作。

4、global，全局變量，嗯，不錯（這算什麽總結嘛）。

python list的+，+=，append，extend