c++ - 为什么 "decltype(i+j)"的结果不是右值引用？

Question

我正在尝试为导致右值的操作提出一个简单的示例。

这个测试用例应该可以工作，但令人惊讶的是(对我而言)，添加两个 int 的结果不是右值(引用)。我在这里错过了什么？

void test(int i, int j)
{
    // this assert should pass, but fails:
    static_assert(std::is_same<decltype(i + j), int&&>(), "i + j should be a rvalue"); 
    // this assert passed, but should fail:
    static_assert(std::is_same<decltype(i + j), int>(), "this assert should fail...");
}

Answer 1

i + j 是 prvalue expression ,

A prvalue ("pure rvalue") expression is an expression that does not have identity and can be moved from.

a + b, a % b, a & b, a << b, and all other built-in arithmetic expressions;

不是 xvalue ,

An xvalue ("expiring value") expression is an expression that has identity and can be moved from.

还有 decltype specifier为纯右值生成 T，而不是 T&&。

a) if the value category of expression is xvalue, then decltype yields T&&;
b) if the value category of expression is lvalue, then decltype yields T&;
c) if the value category of expression is prvalue, then decltype yields T.

您可以通过 std::move:

使其成为 xvalue

static_assert(std::is_same<decltype(std::move(i + j)), int&&>(), "std::move(i + j) is a xvalue then this static_assert won't fail");