C++ 使用 STL 列表，如何将现有列表复制到新列表中
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C++ using STL List, how to copy an existing list into a new list
提问by Ctak
Right now I'm working with a copy constructor for taking a list called val of type char, and I need to take all the elements of a string v that is passed into the copy constructor and put them into the val list.
现在我正在使用一个复制构造函数来获取一个名为 val 的 char 类型的列表，我需要获取传递给复制构造函数的字符串 v 的所有元素并将它们放入 val 列表中。
Public:
LongInt(const string v);
Private:
list<char> val;
So here in the public section of the LongInt class I have a copy constructor which takes the val list and copies the v string into it. Can anyone help me figure out how to do this? Thanks in advance!
所以在 LongInt 类的公共部分，我有一个复制构造函数，它接受 val 列表并将 v 字符串复制到其中。谁能帮我弄清楚如何做到这一点？提前致谢！
采纳答案by Mark B
In your LongIntconstructor just use the iterator, iteratorlistconstructor:
在您的LongInt构造函数中，只需使用iterator, iteratorlist构造函数：
LongInt(const string v) : val(v.begin(), v.end()) { }
LongInt(const string v) : val(v.begin(), v.end()) { }
That being said, have you considered actually using stringor possibly deque<char>to manipulate your sequence rather than list? Depending on your needs, those alternatives might be better.
话虽如此，您是否考虑过实际使用string或可能deque<char>操纵您的序列而不是list？根据您的需要，这些替代方案可能更好。
回答by Timo Geusch
You'll have to iterate over the string and extract the data character by character. Using the std::copyalgorithm should work:
您必须遍历字符串并逐个字符地提取数据。使用该std::copy算法应该有效：
std::copy(v.begin(), v.end(), std::back_inserter(val));
回答by f4.
LongInt::LongInt( const string v ) : val(v.begin(), v.end())
{
}
回答by rubenvb
First, use std::stringif it's a string you're storing. It's a container like any other. If you can't or don't want to store a string, use std::vector. But that would boil down to a less-functional std::stringanyway, so just use std::string.
首先，std::string如果它是您要存储的字符串，请使用。这是一个和其他容器一样的容器。如果您不能或不想存储字符串，请使用std::vector. 但这std::string无论如何都会归结为功能较少，所以只需使用std::string.
For the copying:
对于复制：
std::copy( v.begin(), v.end(), std::back_inserter(val) );
But just use a std::stringif it's a list of chars you're storing.
但是，std::string如果它char是您要存储的s列表，请使用 a 。
