使用迭代器的泛型函数中的 const 正确性

Question

I want to write a generic functions that takes in a sequence, while guaranteeing to not alter said sequence.

template<typename ConstInputIter, typename OutputIter>
OutputIter f(ConstInputIter begin, ConstInputIter end, OutputIter out)
{
  InputIter iter = begin;
  do
  {
    *out++ = some_operation(*iter);
  }while(iter!=end);
  return out;
}

Yet the above example still would take any type as ConstInputIterator , not just const ones. So far, the notion towards being const in it is nominal.

How do I declare the sequence given will not be altered by this function?

Answer 1

Even in C++20, there is no generic way to coerce an iterator over a non- const T into an iterator over a T const . Particular iterators may have a mechanism to do that, and you can use std::cbegin/cend for ranges to get const iterators. But given only an iterator, you are at the mercy of what the user provides.

Applying a C++20 constraint (requiring iter_value_t to be const ) is the wrong thing, as your function should be able to operate on a non- const range.

Answer 2

Since c++17 , you can use std::as_const every time you deference your iterator:

// ...
*out++ = some_operation(std::as_const(*iter));
// ...

This makes sure you never access the elements of the underlying container through a non- const l-value reference.

Note: if you don't have access to c++17 , it's pretty trivial to implement your own version of std::as_const . Just make sure you declare it outside of namespace std .