[英]Copy std::stack into an std::vector
Is the following code guaranteed by the standard to work(assuming st is not empty)? 标准是否保证以下代码可以工作(假设st不为空)?
#include <vector>
#include <stack>
int main()
{
extern std::stack<int, std::vector<int> > st;
int* end = &st.top() + 1;
int* begin = end - st.size();
std::vector<int> stack_contents(begin, end);
}
Yes. 是。
std::stack
is just a container adapter. std::stack
只是一个容器适配器。
You can see that .top()
is actually (§23.3.5.3.1) 你可以看到.top
.top()
实际上是(§23.3.5.3.1)
reference top() { return c.back(); }
Where c
is the container, which in this case is a std::vector
其中
c
是容器,在本例中是std::vector
Which means that your code is basically translated into: 这意味着您的代码基本上被翻译成:
extern std::vector<int> st;
int* end = &st.back() + 1;
int* begin = end - st.size();
std::vector<int> stack_contents(begin, end);
And as std::vector
is guaranteed to be continuous there should be no problem. 并且由于
std::vector
保证是连续的,所以应该没有问题。
However, that does not mean that this is a good idea. 但是,这并不意味着这是一个好主意。 If you need to use "hacks" like this it is generally an indicator of bad design.
如果你需要使用这样的“黑客”,它通常是糟糕设计的一个指标。 You probably want to use
std::vector
from the beginning. 你可能想从头开始使用
std::vector
。
Only std::vector
is guaranteed by C++03 to have contiguous elements ( 23.4.1 ). C ++ 03只保证
std::vector
具有连续元素( 23.4.1 )。 In C++1x this will be extended to std::string
as well ( defect #530 ). 在C ++ 1x中,这也将扩展到
std::string
( 缺陷#530 )。
Yes, it's guaranteed. 是的,这是有保证的。 Vectors are guaranteed to use contiguous storage, so your code will work.
保证向量使用连续存储,因此您的代码将起作用。 It's a bit cludgy though - and if someone changes the underlying container type of the stack, your code will continue to compile without errors, yet the runtime behaviour will be broken.
虽然它有点笨拙 - 如果有人更改了堆栈的基础容器类型,您的代码将继续编译而不会出现错误,但运行时行为将被破坏。
I don't have a reference to the standard to back this up unfortunately, but there aren't many ways in which it could go wrong I guess: 遗憾的是,我没有参考标准来支持这个问题,但我认为没有多少方法可以解决这个问题:
std::vector<int>
as the container type means that the elements must be stored in a std::vector<int>
. std::vector<int>
指定为容器类型意味着元素必须存储在std::vector<int>
。 st.top()
must return a reference to an element in the underlying container (ie an element in the std::vector<int>
. Since the requirements on the container are that it supports back()
, push_back()
and pop_back()
, we can reasonably assume that top()
returns a reference to the last element in the vector. st.top()
必须返回对底层容器中元素的引用(即std::vector<int>
的元素。由于对容器的要求是它支持back()
, push_back()
和pop_back()
,我们可以合理地假设top()
返回对向量中最后一个元素的引用。 end
therefore points to one past the last element. end
指向最后一个元素。 start
therefore points to the beginning. start
指向开头。 Conclusion: Unless the assumption was wrong, it must work. 结论:除非假设是错误的,否则它必须有效。
EDIT: And given the other answer's reference to the standard, the assumption is correct, so it works. 编辑:并给出其他答案的标准参考,假设是正确的,所以它的工作原理。
According to this page , std::stack
uses a container class to store elements. 根据这个页面 ,
std::stack
使用容器类来存储元素。
I guess what you suggest works only if the containter store its elements in a linear way ( std::vector
). 我猜你的建议只有当包含者以线性方式(
std::vector
)存储其元素时才有效。
As a default, std::stack
uses a std::deque
which, as far as I know, doesn't meet this requirement. 默认情况下,
std::stack
使用std::deque
,据我所知,它不符合此要求。 But If you specify a std::vector
as a container class, I can't see a reason why it shoudln't work. 但是如果你将
std::vector
指定为容器类,我就看不出它为什么不起作用的原因了。
Edit: initial statement redacted, the standard actually does provide a full definition for the stack adaptor, nothing left to implentors. 编辑:初始语句编辑,标准实际上确实提供了堆栈适配器的完整定义,没有任何留给实现者。 see top answer.
看到最佳答案。
You want a container that has a push and pop method and allows you to inspect elements anywhere in the container and uses a std::vector
for storage. 您需要一个具有push和pop方法的容器,并允许您检查容器中任何位置的元素,并使用
std::vector
进行存储。 There is such a container in the standard template library 标准模板库中有这样的容器
it is called std::vector
. 它被称为
std::vector
。
Use std::stack
only for bondage purposes 仅将
std::stack
用于绑定目的
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