以下代码不会在C11(也不是C14)中编译.我了解编译器的错误输出,但为什么标准不允许？

//main.cpp

#include <vector>

int main(void)
{
    double a = 3.0;
    double b = 3.0;

    //It works with mere pointers
    const double* ptrToConst = &a;
    /***/ double* ptrToObj   = &a;
//  ptrToObj = ptrToConst; //Illegal : that's understandable…
    ptrToConst = ptrToObj;   //Works

    //But the same doesn't work with vectors to pointers
    std::vector<const double*> ptrsToConst = {&a,&b};
    std::vector</***/ double*> ptrsToObj   = {&a,&b};
//  ptrsToObj = ptrsToConst; //Illegal : that's understandable
    ptrsToConst = ptrsToObj; //Illegal : but why?!
}

错误来自ptrsToConst = ptrsToObj.实际上,似乎不可能复制指针的向量std :: vector< T *>变成一个向量指向常量的std :: vector< const T *>.请注意,在这两种情况下,指针本身不是常量.

为什么这个行为是非法的？
最优雅的工作是什么？

更多细节

如果我通过调用clang –std = c 11 main.cpp进行编译,则会显示以下错误消息：

main.cpp:19:17: error: no viable overloaded '='
    ptrsToConst = ptrsToObj; //Illegal : but why?!
    ~~~~~~~~~~~ ^ ~~~~~~~~~
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.4.0/../../../../include/c++/5.4.0/bits/stl_vector.h:436:7: note: candidate
      function not viable: no known conversion from 'vector<double *,allocator<double *>>' to 'const
      vector<const double *,allocator<const double *>>' for 1st argument
      operator=(const vector& __x);
      ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.4.0/../../../../include/c++/5.4.0/bits/stl_vector.h:448:7: note: candidate
      function not viable: no known conversion from 'vector<double *,allocator<double *>>' to 'vector<const
      double *,allocator<const double *>>' for 1st argument
      operator=(vector&& __x) noexcept(_Alloc_traits::_S_nothrow_move())
      ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.4.0/../../../../include/c++/5.4.0/bits/stl_vector.h:470:7: note: candidate
      function not viable: no known conversion from 'std::vector<double *>' to 'initializer_list<value_type>' (aka
      'initializer_list<const double *>') for 1st argument
      operator=(initializer_list<value_type> __l)
      ^
1 error generated.

与gcc(g)一样尝试产生类似的错误消息.

显然,实现向量的方式不允许我正在尝试执行的操作.但是对于正确性来说,这是一个安全的操作呢？