STL不提供这样的迭代器。但你可以自己编写代码（测试代码）：

The STL does not provide such an iterator. But you could code it yourself (tested that code):

struct null_output_iterator : 
    std::iterator< std::output_iterator_tag,
                   null_output_iterator > {
    /* no-op assignment */
    template<typename T>
    void operator=(T const&) { }

    null_output_iterator & operator++() { 
        return *this; 
    }

    null_output_iterator operator++(int) { 
        return *this;
    }

    null_output_iterator & operator*() { return *this; }
};

使用自身作为运算符的结果不需要任何数据* 。在输出迭代器的要求中不使用 * it = x; 的结果，所以我们可以给它一个返回类型 void 。

It doesn't need any data by using itself as the result of operator*. The result of *it = x; is not used in the output iterator requirements, so we can give it a return type of void.

编辑：让我们进入运算符* 工程。标准在 24.1.2 / 1 中说明了在这两种情况下输出迭代器的要求：

Let's go into how this operator* works. The Standard says in 24.1.2/1 about the requirements of an output iterator that in both these cases:

*it = t;
*it++ = t;

不使用这些表达式的结果。这是什么使这项工作：

That the result of those expressions is not used. That's what makes this work:

null_output_iterator it;
*it; // returns a null_output_iterator& per definition of the `operator*`.
*it = some_value; // returns void per definition of the templated `operator=`. 

现在我们不需要在运算符中返回任何数据* ：我们只是使用迭代器本身。请注意，模板化运算符=不覆盖内置拷贝分配运算符。仍然提供。

Now we don't need to have any data that we return in operator*: We just use the iterator itself. Note that the templated operator= does not overwrite the builtin copy assignment operator. It's still provided.