c++ - 如何告诉auto推断vector 的元素的非引用类型

Question

这是代码:

int main()
{
    std::vector<bool> b(5,false);
    auto b0=b[0];
    cout<<b0;
    b[0]=true;
    cout<<b0;
    std::vector<int> i(5,false);
    auto i0=i[0];
    cout<<i0;
    i[0]=true;
    cout<<i0;

    return 0;
}

0100

变量b0具有引用类型(std::_ Bit_reference)，而i0是常规int。告诉auto推断某些非引用类型(例如bool)的正确语法是什么？

Answer 1

您将获得std::_Bit_reference，因为std::vector具有专用于bool模板的化(即std::vector<bool>)的“可能”节省空间的实现。

作为cppreference says， vector 以位而不是字节存储值:

The manner in which std::vector is made space efficient (as well as whether it is optimized at all) is implementation defined. One potential optimization involves coalescing vector elements such that each element occupies a single bit instead of sizeof(bool) bytes.

正如 templatetypedef所说， auto可以推断出第一件事，这恰好是从位获取 bool(boolean) 值的一种奇特类型。

Exposes class std::vector::reference as a method of accessing individual bits. In particular, objects of this class are returned by operator[] by value.

---

>operator bool() const;
>(until C++11)
>operator bool() const noexcept;
>(since C++11)
>Returns the value of the referenced bit.

它可以隐式转换为 bool(boolean) (AFAIK)，因此您不必担心将其传递给需要 bool参数的函数。