std::list; list_type 为 (char * data, int lenght)

发布于 2024-07-27 20:16:41 字数 199 浏览 5 评论 0原文

我有一些

std::list<char> list_type

现在我必须将列表的内容提供为（char *data，int length）。有没有方便的方法将列表内容呈现为指针和长度？有这样的接口吗？

先感谢您。

原文

I have some

std::list<char> list_type

Now I have to supply contents of the list as (char *data, int length). Is there convenient way to present list contents as pointer and length? Does <vector> has such interface?

Thank you in advance.

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请持续率性 2024-08-03 20:16:42

您可以使用向量来完成此操作，因为它的数据是连续存储的：

std::vector<char> vec;

char* data = &vec[0];
int length = static_cast<int>(vec.size());

对于列表，您必须将数据复制到数组中。幸运的是，这也相当简单：

std::list<char> list:
int length = static_cast<int>(list.size());
char* data = new char[length]; // create the output array
std::copy(list.begin(), list.end(), data); // copy the contents of the list to the output array

当然，您会留下一个动态分配的数组，您必须再次释放它。

You can do it with a vector, because its data is stored contiguously:

std::vector<char> vec;

char* data = &vec[0];
int length = static_cast<int>(vec.size());

For list, you have to copy the data to an array. Luckily, that too is fairly easy:

std::list<char> list:
int length = static_cast<int>(list.size());
char* data = new char[length]; // create the output array
std::copy(list.begin(), list.end(), data); // copy the contents of the list to the output array

Of course, you're then left with a dynamically allocated array you have to free again.

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空城旧梦 2024-08-03 20:16:42

您可以使用向量来完成此操作，而不是使用列表。向量保证是连续的内存块，因此您可以说：

char *data = &list_type[0];
std::vector<char>::size_type length = list_type.size();

You can do this with vector, not with list. A vector is guaranteed to be a contigous chunk of memory so you can say:

char *data = &list_type[0];
std::vector<char>::size_type length = list_type.size();

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你不是我要的菜∠ 2024-08-03 20:16:42

我不知道 std::list，但 std::vector 知道：

std::vector<char> list_type;

...

foo(&list_type[0], list_type.size())

std::string 也可以完成这项工作，但你可能已经知道了。

I don't know about std::list, but std::vector does:

std::vector<char> list_type;

...

foo(&list_type[0], list_type.size())

std::string can do the job too, but you probably already know it.

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孤檠 2024-08-03 20:16:42

您无法使用列表执行此操作，因为列表将其数据保存在列表节点中。但是，您可以使用向量来完成此操作，保证将其数据存储在连续的内存中。您可以使用 &v[0] 或 &*v.begin() 来获取指向其第一个元素的指针：

void f(std::list<char>& list)
{
  std::vector<char> vec(list.begin(),list.end());
  assert(!vec.empty());
  c_api_function(&vec[0],vec.size());
  // assuming you need the result of the call to replace the list's content
  list.assign(vec.begin(),vec.end());
}

请注意，向量将自动释放其第一个元素函数返回时的内存。
还有（至少）两件事值得注意：

向量不能为空。不允许您访问空向量的 v[0]。（也不允许取消引用 v.begin()。）
由于涉及动态分配，因此需要在 std::list 和 std:: 之间来回转换。矢量可能是真正的性能杀手。考虑完全切换到 std::vector 。

You cannot do this with a list, as a list saves its data in list nodes. However, you can do this with a vector, which is guaranteed to store its data in a contiguous piece of memory. You can use either &v[0] or &*v.begin() to get a pointer to its first element:

void f(std::list<char>& list)
{
  std::vector<char> vec(list.begin(),list.end());
  assert(!vec.empty());
  c_api_function(&vec[0],vec.size());
  // assuming you need the result of the call to replace the list's content
  list.assign(vec.begin(),vec.end());
}

Note that the vector will automatically free its memory when the function returns.
There are (at least) two more noteworthy things:

The vector must not be empty. You are not allowed to access v[0] of an empty vector. (Neither are you allowed to dereference v.begin().)
Since dynamic allocation is involved, converting back and forth between std::list and std::vector can be a real performance killer. Consider switching to std::vector altogether.

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