c++ - 使用 decltype 和 auto 推导 const(返回)类型

Question

为什么 decltype推断const来自变量而不是函数？

const int f()
{
    const int i = 1;
    return i;
}
decltype(auto) i_fromFunc = f(); // i_fromFunc is of type 'int'

const int i = 1;
decltype(auto) i_fromVar = i; // i_fromVar is of type 'const int'

---

具有更多推导变体的示例 ( see at compiler explorer )。

const int func()
{
    const int i = 1;
    return i;
}

decltype(auto) funcDecltype()
{
    const int i = 1;
    return i;        
}

auto funcAuto()
{
    const int i = 1;
    return i;
}

void DeduceFromFunc()
{
    decltype(auto) i = func(); // i is of type 'int', similar to auto i = func()
    i = 2;
    decltype(auto) iDT = funcDecltype(); // iDT is of type 'int', similar to auto iDT = funcDecltype()
    iDT = 2;
    decltype(auto) iA = funcAuto(); // iA is of type 'int', similar to auto iA = funcAuto().
    iA = 2;

    // For the sake of completeness:

    const auto ci = func(); // const int
    //ci = 2; // Not mutable.

    //auto& iRef = func(); // non const lvalue of 'int&' can't bind to rvalue of 'int'
    const auto& ciRef = func(); // const int &
    //ciRef = 2; // Not mutable.

    auto&& iRV = func(); // int &&
    iRV = 2;
    const auto&& ciRV = func(); // const int &&
    //ciRV = 2; // Not mutable.
}

const int gi = 1;

void DeduceFromVar()
{
    auto i_fromVar = gi; // i_fromVar is of type 'int'.
    i_fromVar = 2;
    decltype(auto) iDT_fromVar = gi; // iDT_fromVar is of type 'const int'
    //iDT = 2; // Not mutable.

    // For the sake of completeness:
    auto& iRef = gi; // reference to const int
    //iRef = 2; // Not mutable.
    auto&& iRVRef = gi; // rvalue ref to const int
    //iRVRef = 2; // Not mutable.
}

int main()
{
    DeduceFromFunc();
    DeduceFromVar();
}

为什么 iDT_fromVar 的常量性不同于 i_fromVar ？ (decltype(auto) 与 auto 对于变量)

为什么 iDT_fromVar 的常量性不同于 iDT ？ (上面的问题)

为什么 i_fromVar和 i/iDT/iA在 iDT_fromVar 时具有相同的常量和 iDT有没有？

Answer 1

没有非类类型的 const 纯右值。函数 const int f() 基本上等同于 int f()。

C++17中的相关子句是[expr]/6:

If a prvalue initially has the type “cv T ”, where T is a cv-unqualified non-class, non-array type, the type of the expression is adjusted to T prior to any further analysis.

如果使用类类型而不是 int 作为返回类型尝试类似的代码，您会看到不同的行为。