lambda 函数对象中的静态变量如何工作?

Are static variables used in a lambda retained across calls of the function wherein the lambda is used? Or is the function object "created" again each function call?

无用示例:

#include <iostream>
#include <vector>
#include <algorithm>

using std::cout;

void some_function()
{
    std::vector<int> v = {0,1,2,3,4,5};
    std::for_each( v.begin(), v.end(),
         [](const int &i)
         {
             static int calls_to_cout = 0;
             cout << "cout has been called " << calls_to_cout << " times.
"
                  << "	Current int: " << i << "
";
             ++calls_to_cout;
         } );
}

int main()
{
    some_function();
    some_function();
}

这个程序的正确输出是什么?它是否取决于 lambda 是否捕获局部变量的事实?(它肯定会改变函数对象的底层实现，所以它可能会产生影响)是否允许行为不一致?

void some_function()
{
    struct /*unnamed unique*/{
      inline void operator()(int const& i) const{
        static int calls_to_cout = 0;
        cout << "cout has been called " << calls_to_cout << " times.
"
             << "	Current int: " << i << "
";
        ++calls_to_cout;

      }
    } lambda;
    std::vector<int> v = {0,1,2,3,4,5};
    std::for_each( v.begin(), v.end(), lambda);
}