“一切”是一种尊重还是不是？ [英] Is 'everything' a refrence or isn't it?

问题描述

我假设python中的所有内容都是一个参考...


所以，如果我编码：

lst = [1,2 ，3]

for i in lst：

if i == 2：

i = 4

print lst


我虽然会修改lst的内容..（看完之后

''所有''是一个参考。）

因此，为了做到这一点，我需要编写代码：


lst = [1,2,3]

for i在范围（len（lst））：

如果lst [i] == 2：

lst [i] = 4

print lst


我误解了什么吗？

解决方案

KraftDiner写道：

我假设python中的所有东西都是参考...


这是真的。

所以如果我编码：
lst = [1,2,3]


lst是对包含对1,2和3的引用的列表的引用

for i在lst：


i是对列表中每个元素的引用，另一个是另一个。

但这不再是一个参考，列表被取消引用
$ b在获取其内容之前$ b，并且仅返回

返回。

如果i == 2：
i = 4

在这个周期中你现在有我作为4的参考。


ciao - chris


-

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2006-01-04，KraftDiner< bo ******* @ yahoo.com>写道：

我假设python中的所有内容都是一个参考...


它是。

所以，如果我编码：
lst = [1,2,3]
我在lst：
如果i == 2：
i = 4
print lst

我虽然会修改lst的内容..


不。 i = 4没有修改对象。它改变了i。

指向另一个对象（一个整数值为4）。

（读完之后
''所有''是一个参考。）
所以似乎为了做到这一点，我需要编写代码：

lst = [1,2,3]
我在范围内（len（lst） ）：
如果lst [i] == 2：
lst [i] = 4
打印lst

我误解了什么吗？

排序。


你误解了i = 4这句话的含义。意味着。


-

格兰特爱德华兹格兰特哇！我会得到它的！！这可能是我b / b
visi.com意大利女孩朋友的价格。

KraftDiner写道：

我假设python中的所有东西都是一个参考...


它是，尽管它更好从名称和绑定的角度思考。

所以，如果我编码：
lst = [1,2,3]
我在lst：
如果我== 2：
我= 4
打印lst

我虽然会修改lst的内容..（看完之后
''一切''是参考。）


执行此代码时，名称我将被绑定为1,2,3和4.

列表元素被绑定到1,2和3.重新绑定''我'这样，而不是

引用2它现在引用4不会影响第二个
$列表中的b $ b元素引用2.

所以看来为了做到这一点，我需要d如下编码：

lst = [1,2,3]
我在范围内（len（lst））：
如果lst [i] == 2 ：
lst [i] = 4
打印lst

我误解了什么吗？

显然。

I was under the assumption that everything in python was a refrence...

so if I code this:
lst = [1,2,3]
for i in lst:
if i==2:
i = 4
print lst

I though the contents of lst would be modified.. (After reading that
''everything'' is a refrence.)
so it seems that in order to do this I need to code it like:

lst = [1,2,3]
for i in range(len(lst)):
if lst[i] == 2:
lst[i]=4
print lst

Have I misunderstood something?

解决方案

KraftDiner wrote:

I was under the assumption that everything in python was a refrence...
This is true.
so if I code this:
lst = [1,2,3]
lst is a reference to a list that holds references to 1, 2, and 3
for i in lst:
i is a reference to every element in the list, one ofter the other.
But this is not a ref to lst any longer, the list gets dereferenced
before getting at its contents, and only references to that are
returned.
if i==2:
i = 4

You now have i as a reference to 4, during this cycle.

ciao - chris

--
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On 2006-01-04, KraftDiner <bo*******@yahoo.com> wrote:

I was under the assumption that everything in python was a refrence...
It is.
so if I code this:
lst = [1,2,3]
for i in lst:
if i==2:
i = 4
print lst

I though the contents of lst would be modified..
Nope. "i = 4" doesn''t modify the object. It changes "i" to
point to a different object (one that is an integer value 4).
(After reading that
''everything'' is a refrence.)
so it seems that in order to do this I need to code it like:

lst = [1,2,3]
for i in range(len(lst)):
if lst[i] == 2:
lst[i]=4
print lst

Have I misunderstood something?

Sort of.

You''ve misunderstood what the statement "i = 4" means.

--
Grant Edwards grante Yow! I''LL get it!! It''s
at probably a FEW of my
visi.com ITALIAN GIRL-FRIENDS!!

KraftDiner wrote:

I was under the assumption that everything in python was a refrence...
It is, although it is better to think in terms of names and bindings.

so if I code this:
lst = [1,2,3]
for i in lst:
if i==2:
i = 4
print lst

I though the contents of lst would be modified.. (After reading that
''everything'' is a refrence.)
During execution of this code, the name ''i'' is bound to 1, 2, 3 and 4. the
list elements are bound to 1, 2, and 3. Rebinding ''i'' such that instead of
referencing 2 it now references 4 doesn''t affect the fact that the second
element of the list references 2.
so it seems that in order to do this I need to code it like:

lst = [1,2,3]
for i in range(len(lst)):
if lst[i] == 2:
lst[i]=4
print lst

Have I misunderstood something?

Evidently.

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