关于std :: list的问题 [英] A Question about std::list
问题描述
当您遍历列表中的对象列表时。
list< object> mylist;
list< object> :: const_iterator iter;
object ob;
for（iter = mylist.begin（ ）; iter！= mylist.end（）; ++ iter）
{
ob = * iter;
ob.value = 10 ;
}
所以这里我正在遍历列表
但是声明是ob = * iter制作副本我列表中的一个元素？
或者我对mylist中的元素有所了解？
我在列表中复制一个元素似乎效率低下，当
我真正想要的只是一个指向对象的指针或对它的参考，以便
我可以修改它...
我认为iter->值不起作用...
评论？
JustSomeGuy写道：
当您遍历列表中的对象列表时。
list< object> mylist;
list< object> :: const_iterator iter;
object ob;
for（iter = mylist.begin（）; iter！= mylist.end（）; + + iter）
{
ob = * iter;
你得到一份副本;
ob.value = 10;
您可以修改副本。现在，你可能想插入：
* iter = ob;
}
所以我在这里迭代列表
但是声明是ob = * iter在我的列表中制作一个元素的副本吗？
或者我对mylist中的元素有所了解？
这似乎效率低下我在列表中制作一个元素的副本，当我真正想要的是指向对象的指针或对它的引用时，我可以修改它...
是的，那效率很低。
我认为iter->值不起作用......
你实际上没试过，是吗？
最好
Kai-Uwe
< blockquote>
" JustSomeGuy" <无** @ nottelling.com>在消息中写道
新闻：DmtAc.757759
Pk3.612408@pd7tw1no ...当您遍历列表列表中的对象。
list< object> mylist;
list< object> :: const_iterator iter;
object ob;
for（iter = mylist.begin（）; iter！= mylist.end（）; + + iter）
{
ob = * iter;
ob.value = 10;
}
所以我在这里迭代列表
但是声明ob = * iter在我的列表中制作一个元素的副本吗？
或者我对mylist中的元素有所了解吗？
对我来说复制一个元素似乎效率低下列表中的一个元素，当我真正想要的只是一个指向对象的指针或对它的参考，以便
我可以修改它...
我认为iter->值不起作用......
评论？
好奇，如果你想修改列表为什么呢？你使用const_iterator？如果
你想要一个指针或对列表中某个元素的引用，你为什么不用
只声明一个？
list< object> :: iterator iter;
object& ref;
object * ptr;
for（iter = mylist.begin（）; iter！= mylist.end（）; ++ iter）
{
ref = * iter;
ptr =& * iter;
ref.value = 10;
ptr-> value = 10;
但是，正如Kai-Uwe所说，只要''iter-> value = 10''即可使用
迭代器不是const迭代器。
john
When you iterate through a list of objects in a list.
list<object> mylist;
list<object>::const_iterator iter;
object ob;
for (iter=mylist.begin(); iter != mylist.end(); ++iter)
{
ob = *iter;
ob.value = 10;
}
So here I am iterating through the list
but is the statement ob = *iter making a copy of an element in my list?
or am I getting a refrence to an element in mylist?
It seems inefficient to me to make a copy of an element in the list, when
all I really want is a pointer to the object or a refrence to it so that
I can modify it...
I assume that iter->value is not going to work...
Comments?
JustSomeGuy wrote:
When you iterate through a list of objects in a list.
list<object> mylist;
list<object>::const_iterator iter;
object ob;
for (iter=mylist.begin(); iter != mylist.end(); ++iter)
{
ob = *iter;
You get a copy;
ob.value = 10;
You modify the copy. Now, you probably want to insert:
*iter = ob;
}
So here I am iterating through the list
but is the statement ob = *iter making a copy of an element in my list?
or am I getting a refrence to an element in mylist?
It seems inefficient to me to make a copy of an element in the list, when
all I really want is a pointer to the object or a refrence to it so that
I can modify it...
Yep, that would be inefficient.
I assume that iter->value is not going to work...
You did not actually try, did you?
Best
Kai-Uwe
"JustSomeGuy" <no**@nottelling.com> wrote in message
news:DmtAc.757759
Pk3.612408@pd7tw1no...When you iterate through a list of objects in a list.
list<object> mylist;
list<object>::const_iterator iter;
object ob;
for (iter=mylist.begin(); iter != mylist.end(); ++iter)
{
ob = *iter;
ob.value = 10;
}
So here I am iterating through the list
but is the statement ob = *iter making a copy of an element in my list?
or am I getting a refrence to an element in mylist?
It seems inefficient to me to make a copy of an element in the list, when
all I really want is a pointer to the object or a refrence to it so that
I can modify it...
I assume that iter->value is not going to work...
Comments?
Curious, if you want to modify the list why did you use a const_iterator? If
you want a pointer or reference to an element in the list, why didn''t you
just declare one?
list<object>::iterator iter;
object& ref;
object* ptr;
for (iter=mylist.begin(); iter != mylist.end(); ++iter)
{
ref = *iter;
ptr = &*iter;
ref.value = 10;
ptr->value = 10;
But as Kai-Uwe said simply ''iter->value = 10'' works provided you use an
iterator not a const iterator.
john
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