对于 const 成员，如何在没有 operator=() 的情况下 push_back?

How to push_back() to a C++ std::vector without using operator=() for which the default definition violates having const members?

struct Item {
  Item(int value)
    : _value(value) {
  }
  const char _value;
}

vector<Item> items;

items.push_back(Item(3));

I'd like to keep the _value const since it should not change after the object is constructed, so the question is how do I initialize my vector with elements without invoking operator=()?

Here is the basic error the g++ v3.4.6 is giving me:

.../3.4.6/bits/vector.tcc: In member function `Item& Item::operator=(const Item&)':
.../3.4.6/bits/vector.tcc:238:   instantiated from `void std::vector<_Tp, _Alloc>::_M_insert_aux(__gnu_cxx::__normal_iterator<typename _Alloc::pointer, std::vector<_Tp, _Alloc> >, const _Tp&) [with _Tp = Item, _Alloc = std::allocator<Item>]'
.../3.4.6/bits/stl_vector.h:564:   instantiated from `void std::vector<_Tp, _Alloc>::push_back(const _Tp&) [with _Tp = Item, _Alloc = std::allocator<Item>]'
item.cpp:170:   instantiated from here
.../3.4.6/bits/vector.tcc:238: error: non-static const member `const char Item::_value', can't use default assignment operator

解决方案

For std::vector<T> the elements are required to be Assignable. You type is not Assignable. An implementation of. std::vector<T> could avoid insisting on this requirement but this would be a disservice as the resulting code wouldn't be portable.

You can use a std::list<T> instead or change the definition of you type. For example you can create an accessor to only read the value but no setter. Of course, assignment would change the value. The choice thus is to either allow this change or allow putting the objects into a container. You won't get both.